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Integrals 7.8 introduces the concept of definite integrals. Unlike indefinite integrals, a definite integral has a unique numerical value and is evaluated between two fixed limits. If F(x) is an anti-derivative of f(x), then the value of a definite integral is obtained by finding the difference between the values of F(x) at the upper and lower limits.
\displaystyle \int_a^b f(x)\,dx=F(b)-F(a),\quad \text{where}\quad F'(x)=f(x)
The evaluation of definite integrals is based on the Fundamental Theorem of Calculus, which establishes a powerful relationship between differentiation and integration. It provides a simple method for evaluating definite integrals by using an anti-derivative of the given function.
In this exercise, we will learn:
- the meaning and notation of definite integrals.
- the significance of the lower limit and upper limit of integration.
- the Fundamental Theorem of Calculus and its role in evaluating definite integrals.
- how to calculate definite integrals using anti-derivatives.
- how to solve all NCERT Exercise 7.8 questions step by step using standard formulas and properties.
The following terms and symbols are commonly used while evaluating definite integrals.
| Symbol / Term | Meaning |
|---|---|
| \displaystyle \int_a^b f(x)\,dx | Definite integral of f(x) from a to b |
| a | Lower limit of integration |
| b | Upper limit of integration |
| F(x) | An anti-derivative of f(x) |
| \left[F(x)\right]_a^b | Represents F(b)-F(a) |
Key Concepts
1. Definite Integral
A definite integral represents the value of an integral evaluated between two fixed limits. Unlike an indefinite integral, it has a unique numerical value.
\displaystyle \int_a^b f(x)\,dx
Here, a is called the lower limit and b is called the upper limit of integration.
If F(x) is an anti-derivative of f(x), then
\displaystyle \int_a^b f(x)\,dx=F(b)-F(a)
2. Definite Integral and Area
Geometrically, a definite integral represents the area associated with the curve y=f(x) between the vertical lines x=a and x=b.
If x is any point in the interval [a,b], then
\displaystyle A(x)=\int_a^x f(t)\,dt
defines the area function, which gives the area enclosed by the curve from a to the variable point x. As x changes, the enclosed area also changes.
3. Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus establishes the relationship between differentiation and integration.
(i) First Fundamental Theorem
If
\displaystyle A(x)=\int_a^x f(t)\,dt
where f is continuous on [a,b], then
\displaystyle A'(x)=f(x)
This means that the derivative of the area function is the original function.
(ii) Second Fundamental Theorem
If F(x) is an anti-derivative of f(x), then
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
This theorem provides the standard method for evaluating definite integrals.
4. No Constant of Integration
Unlike an indefinite integral, a definite integral does not contain the constant of integration C. This is because the constant cancels while evaluating F(b)-F(a).
Example:
\displaystyle \left[x^2+C\right]_1^3=(9+C)-(1+C)=8
Hence, the value of a definite integral is independent of the constant of integration.
5. Conditions for a Definite Integral
To evaluate a definite integral using the Fundamental Theorem of Calculus, the integrand should be well-defined and continuous throughout the interval of integration.
If the function is not defined or is discontinuous at any point in the interval, the definite integral cannot be evaluated directly using this theorem.
6. Steps for Evaluating a Definite Integral
- Find an anti-derivative F(x) of the given integrand.
- Do not include the constant of integration C.
- Evaluate F(b)-F(a).
- Simplify the result to obtain the final numerical value.
Tip: While evaluating a definite integral, first find an appropriate anti-derivative of the integrand. Then substitute the upper limit first followed by the lower limit. Since the constant of integration cancels automatically, it is not written while evaluating definite integrals.
Let us now solve all the NCERT questions step by step in Integrals 7.8.
Question 1: Definite Integrals 7.8
1. Evaluate \displaystyle \int_{-1}^{1}(x+1)\,dx.
Solution
Let, \displaystyle I=\int_{-1}^{1}(x+1)\,dx
First, consider the corresponding indefinite integral, i.e.
\displaystyle \int(x+1)\,dx=\frac{x^2}{2}+x=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{-1}^{1}
\displaystyle =F(1)-F(-1)
Now, substituting the upper and lower limits respectively in equation (1),
We get, \displaystyle I=\left(\frac{1^2}{2}+1\right)-\left(\frac{(-1)^2}{2}-1\right)
\displaystyle =\left(\frac12+1\right)-\left(\frac12-1\right)
\displaystyle =\frac32+\frac12=2
Hence,
\boxed{\displaystyle \int_{-1}^{1}(x+1)\,dx=2}
Question 2: Definite Integrals 7.8
2. Evaluate \displaystyle \int_{2}^{3}\frac{1}{x}\,dx.
Solution
Let, \displaystyle I=\int_{2}^{3}\frac{1}{x}\,dx
First, consider the corresponding indefinite integral, i.e.
\displaystyle \int\frac{1}{x}\,dx=\log|x|=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{2}^{3}
\displaystyle =F(3)-F(2)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\log|3|-\log|2|
Using the property \displaystyle \log a-\log b=\log\left(\frac{a}{b}\right), we get
\displaystyle I=\log\left(\frac{3}{2}\right)
Hence,
\boxed{\displaystyle \int_{2}^{3}\frac{1}{x}\,dx=\log\left(\frac{3}{2}\right)}
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Question 3: Integrals 7.8
3. Evaluate \displaystyle \int_{1}^{2}(4x^3-5x^2+6x+9)\,dx.
Solution
Let, \displaystyle I=\int_{1}^{2}(4x^3-5x^2+6x+9)\,dx
First, consider the corresponding indefinite integral, i.e.
\displaystyle \int(4x^3-5x^2+6x+9)\,dx=x^4-\frac{5}{3}x^3+3x^2+9x=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{1}^{2}
\displaystyle =F(2)-F(1)
Now, substituting the upper and lower limits respectively in equation (1),
We obtain \displaystyle I=\left(2^4-\frac{5}{3}(2)^3+3(2)^2+9(2)\right)-\left(1^4-\frac{5}{3}(1)^3+3(1)^2+9(1)\right)
\displaystyle =\left(16-\frac{40}{3}+12+18\right)-\left(1-\frac{5}{3}+3+9\right)
\displaystyle =\frac{98}{3}-\frac{34}{3}=\frac{64}{3}
Hence,
\boxed{\displaystyle \int_{1}^{2}(4x^3-5x^2+6x+9)\,dx=\frac{64}{3}}
Question 4: Integrals Exercise 7.8
4. Evaluate \displaystyle \int_{0}^{\pi/4}\sin2x\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/4}\sin2x\,dx
First, considering the corresponding indefinite integral and
Using the standard formula \displaystyle \int\sin ax\,dx=-\frac{\cos ax}{a}+C, we get
\displaystyle \int\sin2x\,dx=-\frac{\cos2x}{2}=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{0}^{\pi/4}
\displaystyle =F\!\left(\frac{\pi}{4}\right)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=-\frac{\cos\left(2\cdot\frac{\pi}{4}\right)}{2}-\left(-\frac{\cos(2\cdot0)}{2}\right)
\displaystyle =-\frac{\cos\frac{\pi}{2}}{2}+\frac{\cos0}{2}
Using the standard values \displaystyle \cos\frac{\pi}{2}=0 and \displaystyle \cos0=1, we get
\displaystyle I=0+\frac12=\frac12
Hence,
\boxed{\displaystyle \int_{0}^{\pi/4}\sin2x\,dx=\frac12}
Question 5: Definite Integrals
5. Evaluate \displaystyle \int_{0}^{\pi/2}\cos2x\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/2}\cos2x\,dx
First, considering the corresponding indefinite integral and
Using the standard formula \displaystyle \int\cos ax\,dx=\frac{\sin ax}{a}+C, we get
\displaystyle \int\cos2x\,dx=\frac{\sin2x}{2}=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{0}^{\pi/2}
\displaystyle =F\!\left(\frac{\pi}{2}\right)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\frac{\sin\left(2\cdot\frac{\pi}{2}\right)}{2}-\frac{\sin(2\cdot0)}{2}
\displaystyle =\frac{\sin\pi}{2}-\frac{\sin0}{2}
Using the standard values \displaystyle \sin\pi=0 and \displaystyle \sin0=0, we get
\displaystyle I=0-0=0
Hence,
\boxed{\displaystyle \int_{0}^{\pi/2}\cos2x\,dx=0}
Important: Definite integrals are evaluated by first finding an anti-derivative of the integrand and then applying the limits of integration. The final answer is always a single numerical value, not a family of functions.
Question 6: Integrals 7.8
6. Evaluate \displaystyle \int_{4}^{5}e^x\,dx.
Solution
Let, \displaystyle I=\int_{4}^{5}e^x\,dx
First, considering the corresponding indefinite integral and
Using the standard formula \displaystyle \int e^x\,dx=e^x+C, we get
\displaystyle \int e^x\,dx=e^x=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{4}^{5}
\displaystyle =F(5)-F(4)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=e^5-e^4
Taking e^4 common, we get
\displaystyle I=e^4(e-1)
Hence,
\boxed{\displaystyle \int_{4}^{5}e^x\,dx=e^4(e-1)}
Question 7: Exercise 7.8
7. Evaluate \displaystyle \int_{0}^{\pi/4}\tan x\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/4}\tan x\,dx
First, considering the corresponding indefinite integral and
Using the standard formula \displaystyle \int\tan x\,dx=\log|\sec x|+C, we get
\displaystyle \int\tan x\,dx=\log|\sec x|=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{0}^{\pi/4}
\displaystyle =F\!\left(\frac{\pi}{4}\right)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\log\left|\sec\frac{\pi}{4}\right|-\log|\sec0|
Using the standard values \displaystyle \sec\frac{\pi}{4}=\sqrt2 and \displaystyle \sec0=1, we get
\displaystyle I=\log\sqrt2-\log1
Since \displaystyle \log1=0, we get
\displaystyle I=\log\sqrt2=\frac12\log2
Hence,
\boxed{\displaystyle \int_{0}^{\pi/4}\tan x\,dx=\frac12\log2}
Question 8: Special Integrals 7.8
8. Evaluate \displaystyle \int_{\pi/6}^{\pi/4}\cosec x\,dx.
Solution
Let, \displaystyle I=\int_{\pi/6}^{\pi/4}\cosec x\,dx
First, considering the corresponding indefinite integral and
Using the standard formula \displaystyle \int\cosec x\,dx=\log|\cosec x-\cot x|+C, we get
\displaystyle \int\cosec x\,dx=\log|\cosec x-\cot x|=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{\pi/6}^{\pi/4}
\displaystyle =F\!\left(\frac{\pi}{4}\right)-F\!\left(\frac{\pi}{6}\right)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\log\left(\cosec\frac{\pi}{4}-\cot\frac{\pi}{4}\right)-\log\left(\cosec\frac{\pi}{6}-\cot\frac{\pi}{6}\right)
Using the standard values \displaystyle \cosec\frac{\pi}{4}=\sqrt2,\ \cot\frac{\pi}{4}=1,\ \cosec\frac{\pi}{6}=2 and \displaystyle \cot\frac{\pi}{6}=\sqrt3, we get
\displaystyle I=\log(\sqrt2-1)-\log(2-\sqrt3)
Using the property \displaystyle \log a-\log b=\log\left(\frac{a}{b}\right), we get
\displaystyle I=\log\left(\frac{\sqrt2-1}{2-\sqrt3}\right)
Hence,
\boxed{\displaystyle \int_{\pi/6}^{\pi/4}\cosec x\,dx=\log\left(\frac{\sqrt2-1}{2-\sqrt3}\right)}
Congratulations on completing NCERT Class 12 Maths Part 1! We have covered all the chapters and exercises with detailed explanations and step-by-step solutions. The journey continues with Part 2, where I’ll keep providing easy-to-follow solutions and concept-based explanations. Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, to help you learn and revise more effectively.
Now, let’s proceed to the next question of Definite Integrals 7.8.
Question 9: Integrals 7.8
9. Evaluate \displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}.
Solution
Let, \displaystyle I=\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}
First, considering the corresponding indefinite integral and
Using the standard formula \displaystyle \int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\!\left(\frac{x}{a}\right)+C, where a=1, we get
\displaystyle \int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{0}^{1}
\displaystyle =F(1)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\sin^{-1}(1)-\sin^{-1}(0)
Using the standard values \displaystyle \sin^{-1}(1)=\frac{\pi}{2} and \displaystyle \sin^{-1}(0)=0, we get
\displaystyle I=\frac{\pi}{2}-0=\frac{\pi}{2}
Hence,
\boxed{\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{2}}
Question 10: Special Integrals 7.8
10. Evaluate \displaystyle \int_{0}^{1}\frac{dx}{1+x^2}.
Solution
Let, \displaystyle I=\int_{0}^{1}\frac{dx}{1+x^2}
First, considering the corresponding indefinite integral and
Using the standard formula \displaystyle \int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right)+C, where a=1, we get
\displaystyle \int\frac{dx}{1+x^2}=\tan^{-1}x=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{0}^{1}
\displaystyle =F(1)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\tan^{-1}(1)-\tan^{-1}(0)
Using the standard values \displaystyle \tan^{-1}(1)=\frac{\pi}{4} and \displaystyle \tan^{-1}(0)=0, we get
\displaystyle I=\frac{\pi}{4}-0=\frac{\pi}{4}
Hence,
\boxed{\displaystyle \int_{0}^{1}\frac{dx}{1+x^2}=\frac{\pi}{4}}
Tip: While evaluating a definite integral, first find the corresponding indefinite integral (anti-derivative). Then apply the Second Fundamental Theorem of Calculus by substituting the upper and lower limits. Since a definite integral gives a fixed value, do not write the constant of integration C in the final answer.
Question 11: Definite Integrals 7.8
11. Evaluate \displaystyle \int_{2}^{3}\frac{dx}{x^2-1}.
Solution
Let, \displaystyle I=\int_{2}^{3}\frac{dx}{x^2-1}
First, consider the corresponding indefinite integral, i.e.
\displaystyle \int\frac{dx}{x^2-1}
Using the standard formula \displaystyle \int\frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C, where a=1, we get
\displaystyle \int\frac{dx}{x^2-1}=\frac12\log\left|\frac{x-1}{x+1}\right|=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{2}^{3}
\displaystyle =F(3)-F(2)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\frac12\log\left(\frac{3-1}{3+1}\right)-\frac12\log\left(\frac{2-1}{2+1}\right)
\displaystyle =\frac12\log\left(\frac12\right)-\frac12\log\left(\frac13\right)
Using the property \displaystyle \log a-\log b=\log\left(\frac{a}{b}\right), we get
\displaystyle I=\frac12\log\left(\frac{\frac12}{\frac13}\right)=\frac12\log\left(\frac32\right)
Hence,
\boxed{\displaystyle \int_{2}^{3}\frac{dx}{x^2-1}=\frac12\log\left(\frac32\right)}
Question 12: Definite Integrals
12. Evaluate \displaystyle \int_{0}^{\pi/2}\cos^2x\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/2}\cos^2x\,dx
First, considering the corresponding indefinite integral and
Using the identity \displaystyle \cos^2x=\frac{1+\cos2x}{2}, we get
\displaystyle \int\cos^2x\,dx=\int\frac{1+\cos2x}{2}\,dx
\displaystyle =\frac12\int dx+\frac12\int\cos2x\,dx
Using the standard integration formulas, we get
\displaystyle \int\cos^2x\,dx=\frac{x}{2}+\frac{\sin2x}{4}=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{0}^{\pi/2}
\displaystyle =F\!\left(\frac{\pi}{2}\right)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\left(\frac{\pi/2}{2}+\frac{\sin\pi}{4}\right)-\left(\frac02+\frac{\sin0}{4}\right)
Using the standard values \displaystyle \sin\pi=0 and \displaystyle \sin0=0, we get
\displaystyle I=\frac{\pi}{4}+0-(0+0)=\frac{\pi}{4}
Hence,
\boxed{\displaystyle \int_{0}^{\pi/2}\cos^2x\,dx=\frac{\pi}{4}}
Question 13: Definite Integrals by Substitution
13. Evaluate \displaystyle \int_{2}^{3}\frac{x}{x^2+1}\,dx.
Solution
Let, \displaystyle I=\int_{2}^{3}\frac{x}{x^2+1}\,dx
First, consider the corresponding indefinite integral and
Put x^2+1=u. Then, 2x\,dx=du or x\,dx=\frac12\,du.
Therefore,
\displaystyle \int\frac{x}{x^2+1}\,dx=\frac12\int\frac{du}{u}
\displaystyle =\frac12\log|u|
i.e. \displaystyle I=\frac12\log(x^2+1)=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{2}^{3}
\displaystyle =F(3)-F(2)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\frac12\log(3^2+1)-\frac12\log(2^2+1)
\displaystyle =\frac12\log10-\frac12\log5
Using the property \displaystyle \log a-\log b=\log\left(\frac{a}{b}\right), we get
\displaystyle I=\frac12\log\left(\frac{10}{5}\right)=\frac12\log2
Hence,
\boxed{\displaystyle \int_{2}^{3}\frac{x}{x^2+1}\,dx=\frac12\log2}
Question 14: Definite Integrals 7.8
14. Evaluate \displaystyle \int_{0}^{1}\frac{2x+3}{5x^2+1}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{1}\frac{2x+3}{5x^2+1}\,dx
First, consider the corresponding indefinite integral and split it as follows:
\displaystyle \int\frac{2x+3}{5x^2+1}\,dx=\int\frac{2x}{5x^2+1}\,dx+3\int\frac{dx}{5x^2+1}
Let, \displaystyle \int\frac{2x+3}{5x^2+1}\,dx=I_1+3I_2\quad\ldots(1)
where \displaystyle I_1=\int\frac{2x}{5x^2+1}\,dx
and \displaystyle I_2=\int\frac{dx}{5x^2+1}\,dx
To evaluate I_1, put 5x^2+1=u. Then, 10x\,dx=du or 2x\,dx=\frac15\,du.
Therefore,
\displaystyle I_1=\frac15\int\frac{du}{u}=\frac15\log|u|=\frac15\log(5x^2+1)
To evaluate I_2, first reduce it to the standard form.
\displaystyle I_2=\int\frac{dx}{5x^2+1}=\frac15\int\frac{dx}{x^2+\frac15}=\frac15\int\frac{dx}{x^2+\left(\frac1{\sqrt5}\right)^2}
Using the standard formula \displaystyle \int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C, where a=\frac1{\sqrt5}, we get
\displaystyle I_2=\frac15\left(\sqrt5\tan^{-1}(\sqrt5\,x)\right)=\frac1{\sqrt5}\tan^{-1}(\sqrt5\,x)
Substituting the values of I_1 and I_2 in equation (1), we get
\displaystyle \int\frac{2x+3}{5x^2+1}\,dx=\frac15\log(5x^2+1)+\frac3{\sqrt5}\tan^{-1}(\sqrt5\,x)=F(x)\quad\ldots(2)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_0^1=F(1)-F(0)
Now, substituting the upper and lower limits respectively in equation (2), we obtain
\displaystyle I=\left(\frac15\log6+\frac3{\sqrt5}\tan^{-1}\sqrt5\right)-\left(\frac15\log1+\frac3{\sqrt5}\tan^{-1}0\right)
Since \displaystyle \log1=0 and \displaystyle \tan^{-1}0=0, we get
\displaystyle I=\frac15\log6+\frac3{\sqrt5}\tan^{-1}\sqrt5
Hence,
\boxed{\displaystyle \int_{0}^{1}\frac{2x+3}{5x^2+1}\,dx=\frac15\log6+\frac3{\sqrt5}\tan^{-1}\sqrt5}
Interesting Fact: In this exercise of Integrals 7.8, whenever substitution is used to find the anti-derivative, the variable is changed back to x before applying the limits. In the next exercise 7.9, you will learn a more efficient method in which the limits of integration are changed along with the variable, eliminating the need to substitute back.
Question 15: Integrals 7.8
15. Evaluate \displaystyle \int_{0}^{1}xe^{x^2}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{1}xe^{x^2}\,dx
First, consider the corresponding indefinite integral, i.e.
\displaystyle \int\xe^{x^2}\,dx
Put x^2=u. Then, 2x\,dx=du or x\,dx=\frac12\,du.
Therefore,
\displaystyle \int xe^{x^2}\,dx=\frac12\int e^u\,du
\displaystyle =\frac12e^u
i.e. \displaystyle I=\frac12e^{x^2}=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_0^1
\displaystyle =F(1)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\frac12e^{1^2}-\frac12e^{0^2}
\displaystyle =\frac12e-\frac12=\frac{e-1}{2}
Hence,
\boxed{\displaystyle \int_{0}^{1}xe^{x^2}\,dx=\frac{e-1}{2}}
Interesting Fact: Unlike an indefinite integral, which represents a family of functions, a definite integral always evaluates to a fixed numerical value. This is why the constant of integration C is not included in its final answer.
Question 16: Definite Integrals by Partial Fractions
16. Evaluate \displaystyle \int_{1}^{2}\frac{5x^2}{x^2+4x+3}\,dx.
Solution
Let, \displaystyle I=\int_{1}^{2}\frac{5x^2}{x^2+4x+3}\,dx
First, consider the corresponding indefinite integral, i.e.
\displaystyle \int\frac{5x^2}{x^2+4x+3}\,dx
Dividing 5x^2 by x^2+4x+3, we get
\displaystyle \frac{5x^2}{x^2+4x+3}=5-\frac{20x+15}{x^2+4x+3}
Using partial fractions, we get
\displaystyle -\frac{20x+15}{(x+1)(x+3)}=\frac{5}{2(x+1)}-\frac{45}{2(x+3)}
(For Long Division process and Partial Fractions, you can please refer to Exercise 7.5, especially Q. 6 and Q. 12)
Therefore,
\displaystyle \int\frac{5x^2}{x^2+4x+3}\,dx=\int\left(5+\frac{5}{2(x+1)}-\frac{45}{2(x+3)}\right)dx
Using the standard integration formulas, we get
\displaystyle \int\frac{5x^2}{x^2+4x+3}\,dx=5x+\frac52\log|x+1|-\frac{45}{2}\log|x+3|=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_1^2
\displaystyle =F(2)-F(1)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\left(10+\frac52\log3-\frac{45}{2}\log5\right)-\left(5+\frac52\log2-\frac{45}{2}\log4\right)
\displaystyle =5+\frac52\log\frac32-\frac{45}{2}\log\frac54
Taking \displaystyle \frac52 common, we get
\displaystyle I=5-\frac52\left(9\log\frac54-\log\frac32\right)
Hence,
\boxed{\displaystyle \int_{1}^{2}\frac{5x^2}{x^2+4x+3}\,dx=5-\frac52\left(9\log\frac54-\log\frac32\right)}
Question 17: Definite Integrals 7.8
17. Evaluate \displaystyle \int_{0}^{\pi/4}(2\sec^2x+x^3+2)\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/4}(2\sec^2x+x^3+2)\,dx
First, considering the corresponding indefinite integral and
Using the properties of integration, we get
\displaystyle \int(2\sec^2x+x^3+2)\,dx=2\int\sec^2x\,dx+\int x^3\,dx+\int2\,dx
Using the standard integration formulas, we get
\displaystyle \int(2\sec^2x+x^3+2)\,dx=2\tan x+\frac{x^4}{4}+2x=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{0}^{\pi/4}
\displaystyle =F\!\left(\frac{\pi}{4}\right)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\left(2\tan\frac{\pi}{4}+\frac{\left(\frac{\pi}{4}\right)^4}{4}+2\cdot\frac{\pi}{4}\right)-\left(2\tan0+\frac{0^4}{4}+2\cdot0\right)
Using the standard values \displaystyle \tan\frac{\pi}{4}=1 and \displaystyle \tan0=0, we get
\displaystyle I=2+\frac{\pi^4}{1024}+\frac{\pi}{2}
Hence,
\boxed{\displaystyle \int_{0}^{\pi/4}(2\sec^2x+x^3+2)\,dx=2+\frac{\pi}{2}+\frac{\pi^4}{1024}}
Question 18: Integration Using Formulas
18. Evaluate \displaystyle \int_{0}^{\pi}\left(\sin^2\frac{x}{2}-\cos^2\frac{x}{2}\right)dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi}\left(\sin^2\frac{x}{2}-\cos^2\frac{x}{2}\right)dx
First, considering the corresponding indefinite integral and
Using the identity \displaystyle \cos2A=\cos^2A-\sin^2A, we get
\displaystyle \sin^2\frac{x}{2}-\cos^2\frac{x}{2}=-\cos x
Therefore,
\displaystyle \int\left(\sin^2\frac{x}{2}-\cos^2\frac{x}{2}\right)dx=\int(-\cos x)\,dx
Using the standard integration formula \displaystyle \int\cos x\,dx=\sin x+C, we get
\displaystyle \int\left(\sin^2\frac{x}{2}-\cos^2\frac{x}{2}\right)dx=-\sin x=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{0}^{\pi}
\displaystyle =F(\pi)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=-\sin\pi-(-\sin0)
Using the standard values \displaystyle \sin\pi=0 and \displaystyle \sin0=0, we get
\displaystyle I=0-0=0
Hence,
\boxed{\displaystyle \int_{0}^{\pi}\left(\sin^2\frac{x}{2}-\cos^2\frac{x}{2}\right)dx=0}
Question 19: Integrals Exercise 7.8
19. Evaluate \displaystyle \int_{0}^{2}\frac{6x+3}{x^2+4}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{2}\frac{6x+3}{x^2+4}\,dx
First, consider the corresponding indefinite integral and split the given integral as
\displaystyle \int\frac{6x+3}{x^2+4}\,dx=\int\frac{6x}{x^2+4}\,dx+3\int\frac{dx}{x^2+4}
For the first integral, put x^2+4=u. Then, 2x\,dx=du or 6x\,dx=3\,du.
Therefore,
\displaystyle \int\frac{6x}{x^2+4}\,dx=3\int\frac{du}{u}=3\log|u|=3\log(x^2+4)
Also, using the standard formula \displaystyle \int\frac{dx}{a^2+x^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C, where a=2, in the second integral, we get
\displaystyle \int\frac{dx}{x^2+4}=\frac12\tan^{-1}\left(\frac{x}{2}\right)
Hence,
\displaystyle \int\frac{6x+3}{x^2+4}\,dx=3\log(x^2+4)+\frac32\tan^{-1}\left(\frac{x}{2}\right)=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{0}^{2}
\displaystyle =F(2)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\left(3\log8+\frac32\tan^{-1}1\right)-\left(3\log4+\frac32\tan^{-1}0\right)
Using the standard values \displaystyle \tan^{-1}1=\frac{\pi}{4} and \displaystyle \tan^{-1}0=0, we get
\displaystyle I=3(\log8-\log4)+\frac32\cdot\frac{\pi}{4}
Using the property \displaystyle \log a-\log b=\log\left(\frac{a}{b}\right), we get
\displaystyle I=3\log2+\frac{3\pi}{8}
Hence,
\boxed{\displaystyle \int_{0}^{2}\frac{6x+3}{x^2+4}\,dx=3\log2+\frac{3\pi}{8}}
Question 20: Definite Integrals 7.8
20. Evaluate \displaystyle \int_{0}^{1}\left(xe^x+\sin\frac{\pi x}{4}\right)dx.
Solution
Let, \displaystyle I=\int_{0}^{1}\left(xe^x+\sin\frac{\pi x}{4}\right)dx
First, considering the corresponding indefinite integral and
Using the properties of integration, we get
\displaystyle \int\left(xe^x+\sin\frac{\pi x}{4}\right)dx=\int xe^x\,dx+\int\sin\frac{\pi x}{4}\,dx
To evaluate \displaystyle \int xe^x\,dx, we use the formula for integration by parts,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
Here, u=x and v=e^x. Therefore,
We get \displaystyle \int xe^x\,dx=x\int e^x\,dx-\int\left(\frac{d}{dx}(x)\int e^x\,dx\right)dx
\displaystyle =xe^x-\int e^x\,dx
\displaystyle =xe^x-e^x=(x-1)e^x
Also, using the standard formula \displaystyle \int\sin(ax)\,dx=-\frac{\cos(ax)}{a}+C, where a=\frac{\pi}{4}, we get
\displaystyle \int\sin\frac{\pi x}{4}\,dx=-\frac{4}{\pi}\cos\frac{\pi x}{4}
Hence,
\displaystyle \int\left(xe^x+\sin\frac{\pi x}{4}\right)dx=(x-1)e^x-\frac{4}{\pi}\cos\frac{\pi x}{4}=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{0}^{1}
\displaystyle =F(1)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\left((1-1)e-\frac{4}{\pi}\cos\frac{\pi}{4}\right)-\left((0-1)e^0-\frac{4}{\pi}\cos0\right)
Using the standard values \displaystyle e^0=1, \displaystyle \cos\frac{\pi}{4}=\frac{1}{\sqrt2} and \displaystyle \cos0=1, we get
\displaystyle I=-\frac{4}{\pi\sqrt2}-(-1-\frac{4}{\pi})
\displaystyle =1+\frac{4}{\pi}-\frac{2\sqrt2}{\pi}
Hence,
\boxed{\displaystyle \int_{0}^{1}\left(xe^x+\sin\frac{\pi x}{4}\right)dx=1+\frac{4-2\sqrt2}{\pi}}
Question 21: Integrals 7.8 – MCQ
21. Choose the correct answer.
\displaystyle \int_{1}^{\sqrt3}\frac{dx}{1+x^2} equals
- (A) \displaystyle \frac{\pi}{3}
- (B) \displaystyle \frac{2\pi}{3}
- (C) \displaystyle \frac{\pi}{6}
- (D) \displaystyle \frac{\pi}{12}
Solution
Let, \displaystyle I=\int_{1}^{\sqrt3}\frac{dx}{1+x^2}
First, considering the corresponding indefinite integral and
Using the standard formula \displaystyle \int\frac{dx}{a^2+x^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C, where a=1, we get
\displaystyle \int\frac{dx}{1+x^2}=\tan^{-1}x=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{1}^{\sqrt3}=F(\sqrt3)-F(1)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\tan^{-1}(\sqrt3)-\tan^{-1}(1)
Using the standard values \displaystyle \tan^{-1}(\sqrt3)=\frac{\pi}{3} and \displaystyle \tan^{-1}(1)=\frac{\pi}{4}, we get
\displaystyle I=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}
✅️ Hence, the correct option is (D).
Question 22: Definite Integrals 7.8 – MCQ
22. Choose the correct answer.
\displaystyle \int_{0}^{2/3}\frac{dx}{4+9x^2} equals
- (A) \displaystyle \frac{\pi}{6}
- (B) \displaystyle \frac{\pi}{12}
- (C) \displaystyle \frac{\pi}{24}
- (D) \displaystyle \frac{\pi}{4}
Solution
Let, \displaystyle I=\int_{0}^{2/3}\frac{dx}{4+9x^2}
First, considering the corresponding indefinite integral and rewriting
\displaystyle \int\frac{dx}{4+9x^2}=\int\frac{dx}{2^2+(3x)^2}
Put 3x=t. Then, 3\,dx=dt or dx=\frac13\,dt.
Therefore,
\displaystyle \int\frac{dx}{4+9x^2}=\frac13\int\frac{dt}{2^2+t^2}
Using the standard formula \displaystyle \int\frac{dx}{a^2+x^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C, where a=2, we get
\displaystyle \int\frac{dx}{4+9x^2}=\frac13\left(\frac12\tan^{-1}\frac{t}{2}\right)=\frac16\tan^{-1}\frac{3x}{2}=F(x)\quad\ldots(1)
By the Second Fundamental Theorem of Calculus, we have
\displaystyle \int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a)
where F(x) is an anti-derivative of f(x).
Therefore,
\displaystyle I=\left[F(x)\right]_{0}^{2/3}=F\!\left(\frac23\right)-F(0)
Now, substituting the upper and lower limits respectively in equation (1), we obtain
\displaystyle I=\frac16\tan^{-1}\left(\frac{3(2/3)}{2}\right)-\frac16\tan^{-1}(0)
\displaystyle =\frac16\tan^{-1}(1)-\frac16\tan^{-1}(0)
Using the standard values \displaystyle \tan^{-1}(1)=\frac{\pi}{4} and \displaystyle \tan^{-1}(0)=0, we get
\displaystyle I=\frac16\cdot\frac{\pi}{4}=\frac{\pi}{24}
✅️ Hence, the correct option is (C).
Common Mistakes to Avoid
- Adding the constant of integration: A definite integral has a fixed value. Do not write the constant of integration C while evaluating a definite integral.
- Applying the Second Fundamental Theorem incorrectly: First find the corresponding anti-derivative F(x), then evaluate \displaystyle F(b)-F(a). Do not substitute the limits directly into the integrand.
- Interchanging the limits: Always subtract the value at the lower limit from the value at the upper limit, i.e., \displaystyle \int_a^b f(x)\,dx=F(b)-F(a).
- Making mistakes while substituting the limits: Substitute the upper and lower limits carefully in the complete anti-derivative F(x), especially when it contains brackets, fractions or trigonometric functions.
- Ignoring the domain of the integrand: Ensure that the integrand is well-defined and continuous throughout the interval of integration before applying the Second Fundamental Theorem.
- Forgetting to simplify the integrand first: If the integrand can be simplified using identities, partial fractions or algebraic manipulation, do so before finding the anti-derivative.
- Making errors in standard trigonometric values: Evaluate values such as \sin0, \cos\frac{\pi}{2}, \tan^{-1}(1) and \sin^{-1}(1) carefully while substituting the limits.
- Using incorrect logarithmic properties: While simplifying expressions, remember that \displaystyle \log a-\log b=\log\left(\frac{a}{b}\right). Avoid combining logarithms incorrectly.
- Changing the limits after substitution: In this exercise, substitution is used only to obtain the corresponding indefinite integral. The original limits are then substituted in F(x). Changing the limits will be studied in the next exercise.
- Not writing the final answer in its simplest form: Simplify the result by using logarithmic properties, trigonometric identities or algebraic simplification wherever possible before writing the final answer.
Continue Learning
After completing Integrals 7.8, you should now be familiar with the basic concept of definite integrals. You have also learned how to evaluate definite integrals by first finding the corresponding anti-derivative and then applying the Second Fundamental Theorem of Calculus.
To strengthen your understanding further, make sure that you revise:
- The meaning and notation of a definite integral \displaystyle \int_a^b f(x)\,dx
- The significance of the lower limit and upper limit of integration
- The Second Fundamental Theorem of Calculus and the relation \displaystyle \int_a^b f(x)\,dx=F(b)-F(a)
- Finding the corresponding anti-derivative before substituting the limits
- Evaluating definite integrals using standard integration formulas and techniques such as substitution, partial fractions and integration by parts wherever required
- Using standard trigonometric values and logarithmic properties while simplifying the final answer
- Remembering that a definite integral has a fixed value, so the constant of integration C is not written
- Substituting the upper and lower limits carefully into the complete anti-derivative F(x)
- Writing the final answer in its simplest form after applying the required algebraic or trigonometric simplifications
- Recognising that changing the limits after substitution is a different method, which will be studied in the next exercise
Explore More
Definite integrals become much easier with regular practice. While solving questions, first find the corresponding anti-derivative carefully, then apply the Second Fundamental Theorem of Calculus to evaluate the integral. Always substitute the upper and lower limits correctly, simplify the final answer using the appropriate algebraic or trigonometric properties, and remember that a definite integral has a fixed value, so the constant of integration C is not required. With consistent practice, you’ll be able to evaluate definite integrals accurately and confidently.
All the best and keep learning 👍
