Application of Integrals Miscellaneous NCERT Solutions

Example. Find the area of the region bounded by the line y=3x+2, the x-axis and the ordinates x=-1 and x=1.

Solution

Think First… Before forming the definite integral, draw a pre-rough sketch to understand the position of the graph with respect to the x-axis. Since the line crosses the x-axis within the given interval, the required area must be divided into two parts.

Step 1: Analyse the Given Line

The given equation

y=3x+2

represents a straight line.

To find the point where the line cuts the x-axis, put y=0.

3x+2=0

or 3x=-2

\displaystyle \Rightarrow x=-\frac23

Thus, the line intersects the x-axis at the point \left(-\frac23,0\right).

The rough sketch above showing the line, the x-axis, the ordinates x=-1 and x=1, and the point \left(-\frac23,0\right).

From the graph, we observe that:

• The line lies below the x-axis for -1\le x\le-\dfrac23.
• The line lies above the x-axis for -\dfrac23\le x\le1.

Hence, the required area is divided into two regions:

• A_1 = Area of region ACBA (below the x-axis)
• A_2 = Area of region ADEA (above the x-axis)

Therefore, the required area is given by

\displaystyle \text{Required Area}=|A_1|+A_2

where, for the region ACBA,

\displaystyle A_1=\int_{-1}^{-\frac23}(\text{y of the line})\,dx

Since the equation of the line is y=3x+2, we get

\displaystyle A_1=\int_{-1}^{-\frac23}(3x+2)\,dx

and for the region ADEA,

\displaystyle A_2=\int_{-\frac23}^{1}(\text{y of the line})\,dx

Since the equation of the line is y=3x+2, we get

\displaystyle A_2=\int_{-\frac23}^{1}(3x+2)\,dx

Now evaluate A_1 and A_2 separately.

Step 2: Evaluate the Two Integrals

First, evaluate A_1.

\displaystyle A_1=\int_{-1}^{-\frac23}(3x+2)\,dx

Integrating, we get

We get, \displaystyle A_1=\left[\frac32x^2+2x\right]_{-1}^{-\frac23}

\displaystyle =\left[\left\{\frac32\left(-\frac23\right)^2+2\left(-\frac23\right)\right\}-\left\{\frac32(-1)^2+2(-1)\right\}\right]

\displaystyle =\left[\left(\frac32\times\frac49-\frac43\right)-\left(\frac32-2\right)\right]

Simplifying, we get

\displaystyle A_1=\left[\left(\frac23-\frac43\right)-\left(-\frac12\right)\right]

\displaystyle =-\frac23+\frac12=-\frac16

Therefore, \boxed{\displaystyle A_1=-\frac16}

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Now, evaluate A_2.

\displaystyle A_2=\int_{-\frac23}^{1}(3x+2)\,dx

Integrating,

We get, \displaystyle A_2=\left[\frac32x^2+2x\right]_{-\frac23}^{1}

\displaystyle =\left[\left\{\frac32(1)^2+2(1)\right\}-\left\{\frac32\left(-\frac23\right)^2+2\left(-\frac23\right)\right\}\right]

\displaystyle =\left[\left(\frac32+2\right)-\left(\frac32\times\frac49-\frac43\right)\right]

Simplifying,

We get, \displaystyle A_2=\left[\frac72-\left(\frac23-\frac43\right)\right]

\displaystyle =\frac72+\frac23

\displaystyle =\frac{25}{6}

Therefore, \boxed{\displaystyle A_2=\frac{25}{6}}

Step 3: Find the Required Area

From Step 2, we have

\displaystyle A_1=-\frac16\qquad\text{and}\qquad A_2=\frac{25}{6}

Since A_1 represents the signed area below the x-axis, its value is negative. Therefore, while finding the required area, we take its absolute value.

\displaystyle \text{Required Area}=|A_1|+A_2

\displaystyle =\left|-\frac16\right|+\frac{25}{6}=\frac16+\frac{25}{6}

i.e. \displaystyle Area=\frac{26}{6}=\frac{13}{3}

\therefore \boxed{\displaystyle \text{Required Area}=\frac{13}{3}\text{ square units}}

Maths Better Observation: A definite integral represents the signed area. Portions of the graph above the x-axis contribute positively, while portions below the x-axis contribute negatively. Therefore, whenever the graph crosses the x-axis within the given interval, first divide the required region into separate parts and then add their positive areas.

1 (i). Find the area under the curve y=x^2, the lines x=1, x=2 and the x-axis.

Solution

The given curve

y=x^2

is an upward opening parabola with vertex at (0,0). Since the interval 1\le x\le2 lies entirely in the first quadrant, the curve remains above the x-axis throughout the given interval.

The rough sketch above is showing the parabola, the x-axis and the ordinates x=1 and x=2. And the shaded region enclosed by them is the required area.

Let A be the required area.

Then, \displaystyle A=\int_{1}^{2}(\text{y of the curve})\,dx

Since y=x^2, we get

\displaystyle A=\int_{1}^{2}x^2\,dx

Integrating,

We get, \displaystyle A=\left[\frac{x^3}{3}\right]_1^2

\displaystyle =\left[\frac{2^3}{3}-\frac{1^3}{3}\right]

i.e. \displaystyle =\frac83-\frac13=\frac73

Therefore, \boxed{\displaystyle A=\frac73\text{ square units}}

Maths Better Observation: Since the entire curve lies above the x-axis over the given interval, the definite integral itself gives the required area. No absolute value is required.

Q. 1 Part (ii)

1 (ii). Find the area under the curve y=x^4, the lines x=1, x=5 and the x-axis.

Solution

The given curve

y=x^4

is an upward opening curve that resembles the graph of y=x^2, but is narrower, with its two arms closer to each other. It lies entirely above the x-axis over the given interval 1\le x\le5 as shown in the graph below:

Therefore, the required area is equal to the corresponding definite integral.

Let A be the required area (shaded region as shown in the graph)

Then, \displaystyle A=\int_{1}^{5}(\text{y of the curve})\,dx

Since y=x^4, we get

\displaystyle A=\int_{1}^{5}x^4\,dx

Integrating,

We get, \displaystyle A=\left[\frac{x^5}{5}\right]_1^5

\displaystyle =\left[\frac{5^5}{5}-\frac{1^5}{5}\right]

\displaystyle =\left[\frac{3125}{5}-\frac15\right]

or \displaystyle A=625-\frac15=624\frac45

\displaystyle =624.8

Therefore, \boxed{\displaystyle A=624.8\text{ square units}}

2. Sketch the graph of y=|x+3| and evaluate \displaystyle \int_{-6}^{0}|x+3|\,dx.

Solution

The graph of y=|x+3| is a V-shaped graph with vertex at (-3,0). The graph lies above the x-axis throughout the given interval -6\le x\le0.

Since the graph changes its equation at x=-3, divides the required region into two parts.

Let A_1 be the area from x=-6 to x=-3 and A_2 be the area from x=-3 to x=0.

Then, as shown in the graph, the shaded region gives:

\displaystyle \text{Required Area}=A_1+A_2

For the first region,

\displaystyle A_1=\int_{-6}^{-3}(\text{y of the graph})\,dx

Since

|x+3|=\begin{cases} -(x+3), & x<-3 \\ \\ x+3, & x\geq-3 \end{cases}

we get

\displaystyle A_1=\int_{-6}^{-3}-(x+3)\,dx

Integrating,

We get, \displaystyle A_1=\left[-\frac{x^2}{2}-3x\right]_{-6}^{-3}

\displaystyle =\left[\left(-\frac{(-3)^2}{2}-3(-3)\right)-\left(-\frac{(-6)^2}{2}-3(-6)\right)\right]

\displaystyle =\left[\left(-\frac92+9\right)-(-18+18)\right]

i.e. \displaystyle A_1=\frac92

Similarly, for the second region,

\displaystyle A_2=\int_{-3}^{0}(\text{y of the graph})\,dx

Since |x+3|=x+3 for x\geq-3,

We get, \displaystyle A_2=\int_{-3}^{0}(x+3)\,dx

\displaystyle =\left[\frac{x^2}{2}+3x\right]_{-3}^{0}

\displaystyle =\left[(0+0)-\left(\frac{(-3)^2}{2}+3(-3)\right)\right]

i.e. \displaystyle A_2=-\left(\frac92-9\right)=\frac92

Therefore,

\displaystyle \text{Required Area}=A_1+A_2=\frac92+\frac92=9

\boxed{\displaystyle \int_{-6}^{0}|x+3|\,dx=9}

3. Find the area bounded by the curve y=\sin x between x=0 and x=2\pi.

Solution

The graph of y=\sin x intersects the x-axis at x=0,\ \pi,\ 2\pi. It lies above the x-axis over the interval 0\le x\le\pi and below the x-axis over the interval \pi\le x\le2\pi.

The rough sketch below is showing the graph of y=\sin x between x=0 and x=2\pi and the shaded region is the required area.

Let A_1 be the area between x=0 and x=\pi, and A_2 be the area between x=\pi and x=2\pi.

Then, \displaystyle \text{Required Area}=A_1+|A_2|

For the first region,

\displaystyle A_1=\int_{0}^{\pi}(\text{y of the curve})\,dx

Since y=\sin x,

We get, \displaystyle A_1=\int_{0}^{\pi}\sin x\,dx

\displaystyle =\left[-\cos x\right]_{0}^{\pi}

\displaystyle =[-\cos\pi]-[-\cos0]

i.e. \displaystyle A_1=[-(-1)]-[-1]=1+1=2

Similarly, for the second region,

\displaystyle A_2=\int_{\pi}^{2\pi}(\text{y of the curve})\,dx

Since y=\sin x,

We get, \displaystyle A_2=\int_{\pi}^{2\pi}\sin x\,dx

\displaystyle =\left[-\cos x\right]_{\pi}^{2\pi}

\displaystyle =[-\cos2\pi]-[-\cos\pi]

i.e. \displaystyle A_2=[-1]-[1]=-2

Therefore, \displaystyle \text{Required Area}=2+|-2|

\displaystyle =2+2=4

\boxed{\displaystyle \text{Required Area}=4\text{ square units}}

4. Choose the correct answer.

The area bounded by the curve y=x^3, the x-axis and the ordinates x=-2 and x=1 is

• (A) -9
• (B) \displaystyle -\frac{15}{4}
• (C) \displaystyle \frac{15}{4}
• (D) \displaystyle \frac{17}{4}

Solution

Think First… The curve y=x^3 cuts the x-axis at the origin. Therefore, divide the required region into two parts, one below the x-axis and the other above it, and then add their areas.

The curve y=x^3 intersects the x-axis at x=0. It lies below the x-axis for -2\le x\le0 and above the x-axis for 0\le x\le1, as shown in the graph below:

Let A_1 and A_2 denote the signed areas of the two regions.

Then, \displaystyle \text{Required Area}=|A_1|+A_2

For the first region,

\displaystyle A_1=\int_{-2}^{0}(\text{y of the curve})\,dx

Since y=x^3,

We get, \displaystyle A_1=\int_{-2}^{0}x^3\,dx

\displaystyle =\left[\frac{x^4}{4}\right]_{-2}^{0}

\displaystyle =0-\frac{(-2)^4}{4}=0-\frac{16}{4}=-4

Similarly, for the second region,

\displaystyle A_2=\int_{0}^{1}(\text{y of the curve})\,dx

Since y=x^3,

We get, \displaystyle A_2=\int_{0}^{1}x^3\,dx

\displaystyle =\left[\frac{x^4}{4}\right]_{0}^{1}

\displaystyle =\frac14-0=\frac14

Therefore,

\displaystyle \text{Required Area}=|-4|+\frac14=4+\frac14=\frac{17}{4}

✅️ Hence, the correct answer is (D).

5. Choose the correct answer.

The area bounded by the curve y=x|x|, the x-axis and the ordinates x=-1 and x=1 is given by

• (A) 0
• (B) \displaystyle \frac13
• (C) \displaystyle \frac23
• (D) \displaystyle \frac43

Solution

Think First… The function x|x| changes its expression at x=0. First write it as a piecewise function, then evaluate the areas on the two intervals separately.

Since

x|x|=\begin{cases} -x^2, & x<0 \\ \\ x^2, & x\geq0 \end{cases}

the curve lies below the x-axis for -1\le x\le0 and above the x-axis for 0\le x\le1, as shown in the graph:

Let A_1 and A_2 denote the signed areas of the two regions.

Then, \displaystyle \text{Required Area}=|A_1|+A_2

For the first region,

\displaystyle A_1=\int_{-1}^{0}(\text{y of the curve})\,dx

Since y=-x^2,

We get, \displaystyle A_1=\int_{-1}^{0}(-x^2)\,dx

\displaystyle =\left[-\frac{x^3}{3}\right]_{-1}^{0}

\displaystyle =0-\frac13=-\frac13

For the second region,

\displaystyle A_2=\int_{0}^{1}(\text{y of the curve})\,dx

Since y=x^2,

We get, \displaystyle A_2=\int_{0}^{1}x^2\,dx

\displaystyle =\left[\frac{x^3}{3}\right]_{0}^{1}

\displaystyle =\frac13-0=\frac13

Therefore,

\displaystyle \text{Required Area}=\left|-\frac13\right|+\frac13=\frac13+\frac13=\frac23

✅️ Hence, the correct answer is (C).