Application of Integrals MCQ Quiz - Objective Question with Answer for Application of Integrals - Download Free PDF

Concept:

The area under the curve y = f(x) between x = a and x = b,is given by,  Area = ${\int }_{\mathrm{x}=\mathrm{a}}^{\mathrm{x}=\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}$

$\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}$

Calculation:

Here, we have to find the area of the region bounded by the curves y = $\frac{{\mathrm{x}}^2}{2}$, the line x = 2, x  = 0 and the x - axis

So, the area enclosed by the given curves = ${\int }_0^2\frac{{\mathrm{x}}^2}{2}\mathrm{d}\mathrm{x}$

As we know that, $\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}$

$={\int }_0^2\frac{{\mathrm{x}}^2}{2}\mathrm{d}\mathrm{x}={\left[\frac{{\mathrm{x}}^3}{6}\right]}_0^2$

$=\frac{1}{6}\left(8-0\right)=\frac{4}{3}\mathrm{s}\mathrm{q}.\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$

Hence, option 4 is the correct answer.

Calculation

Area = ${\int }_0^{\pi }|\cos x|dx$

cos x is positive from 0 to π/2 and negative from π/2 to π.

So, we split the integral:

Area = ${\int }_0^{\frac{\pi }{2}}\cos xdx+{\int }_{\frac{\pi }{2}}^{\pi }(-\cos x)dx$

Area = ${[\sin x]}_0^{\frac{\pi }{2}}-{[\sin x]}_{\frac{\pi }{2}}^{\pi }$

Area = $(\sin \frac{\pi }{2}-\sin 0)-(\sin \pi -\sin \frac{\pi }{2})$

Area = $(1-0)-(0-1)$

Area = $1-(-1)$

Area = $1+1$

Area = $2$

∴ The area bounded by the curve y = cos x, x = 0 and x = π is 2.

Hence option 1 is correct

Calculation:

Find the equation of line first intercept of y is 2

equation becomes:

y = x + 2

If intercept of y is n

Equation becomes

y = x + n

I = $\underset{1}{\overset{3}{\int }}{y}^{2}dx$

I = $\underset{1}{\overset{3}{\int }}{\left(x+n\right)}^{2}dx$

I = $\underset{1}{\overset{3}{\int }}\left[{\mathrm{x}}^2+{\mathrm{n}}^2+2\mathrm{x}\mathrm{n}\right]\mathrm{d}\mathrm{x}$

n = 2

I = $\underset{1}{\overset{3}{\int }}\left({\mathrm{x}}^2+4+4\mathrm{x}\right)\mathrm{d}\mathrm{x}$

I = ${\left[\frac{{\mathrm{x}}^2}{3}+4x+\frac{4{\mathrm{x}}^2}{3}\right]}_1^3$

$\begin{array}{l}\mathrm{I}=\left[\frac{3^3}{3}+12+18-\frac{1}{3}-4-2\right]\\ \mathrm{I}=33-\frac{1}{3}\end{array}$

Calculation

$\cos x>\sin x,\mathrm{\forall }x\in (0,\frac{\pi }{4})$ and $\cos x<\sin x,\mathrm{\forall }x\in (\frac{\pi }{4},\frac{\pi }{2})$

$y_1=\sin x+\cos x$

$y_2=|\cos x-\sin x|$

$\Rightarrow \text{Area}$

${\int }_0^{\pi /2}(y_1-y_2)dx$

${\int }_0^{\pi /4}((\sin x+\cos x)-(\cos x-\sin x))dx+{\int }_{\pi /4}^{\pi /2}((\sin x+\cos x)-(\sin x-\cos x))dx$

$=4-2\sqrt{2}$

Hence option 2 is correct

Calculation

Given:

Parabola: y² = 4x

Line: x = 3

Since y² = 4x, then y = $2\sqrt{x}$ (in the first quadrant, y > 0)

Required Area = 2 × (Area of the region OCAO)

⇒ Area = $2{\int }_0^{3}ydx$

⇒ Area = $2{\int }_0^{3}2\sqrt{x}dx$

⇒ Area = $4{\int }_0^3{x}^{\frac{1}{2}}dx$

⇒ Area = $4{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^3$

⇒ Area = $4\times \frac{2}{3}[(3{)}^{\frac{3}{2}}-0]$

⇒ Area = $\frac{8}{3}(3\sqrt{3})$

∴ Required area = $8\sqrt{3}$ sq. units.

Hence option 2 is correct

Concept:

The area under the curve y = f(x) between x = a and x = b, is given by:

Area = ${\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{y}\mathrm{d}\mathrm{x}$

Similarly, the area under the curve y = f(x) between y = a and y = b, is given by:

Area = ${\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}\mathrm{d}\mathrm{y}$

Calculation:

Here, 

x² = y  and line y = 1 cut the parabola

∴ x² = 1

⇒ x = 1 and -1

$\text{Area =}{\int }_{-1}^{1}ydx$

Here, the area is symmetric about the y-axis, we can find the area on one side and then multiply it by 2, we will get the area,

${\text{Area}}_1={\int }_0^{1}ydx$

${\text{Area}}_1={\int }_0^1{x}^{2}dx$

$={\left[\frac{{\mathrm{x}}^3}{3}\right]}_0^1=\frac{1}{3}$

This area is between y = x² and the positive x-axis.

To get the area of the shaded region, we have to subtract this area from the area of square i.e.

$(1\times 1)-\frac{1}{3}=\frac{2}{3}$

$TotalArea=2\times \frac{2}{3}=\frac{4}{3}$ square units.

Concept: 

$\int \surd a^2-x^{2}dx=\frac{x}{2}\surd x^2-a^2+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}+c$ 

Function y = √f(x) is defined for f(x) ≥ 0. Therefore y can not be negative.

Calculation:

Given: 

y = $\surd 16-{\mathrm{x}}^2$ and x-axis

At x-axis, y will be zero

y = $\surd 16-{\mathrm{x}}^2$

⇒ 0 = $\surd 16-{\mathrm{x}}^2$

⇒ 16 - x² = 0

⇒ x2 = 16

∴ x = ± 4

So, the intersection points are (4, 0) and (−4, 0)

Since the curve is y = $\surd 16-{\mathrm{x}}^2$

So, y ≥ o [always]

So, we will take the circular part which is above the x-axis

Area of the curve, A $={\int }_{-4}^4\surd 16-{\mathrm{x}}^2\mathrm{d}\mathrm{x}$

We know that,

$\int \surd a^2-x^{2}dx=\frac{x}{2}\surd x^2-a^2+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}+c$

= $[\frac{\mathrm{x}}{2}\surd (4^2-{\mathrm{x}}^2)+\frac{16}{2}\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\frac{\mathrm{x}}{4}{]}_{-4}^4$ 

= $[\frac{\mathrm{x}}{2}\surd (4^2-4^2)+\frac{16}{2}\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\frac{4}{4}]-[\frac{\mathrm{x}}{2}\surd (4^2-(-4{)}^2)+\frac{16}{2}\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\frac{4}{-4})]$

= 8 sin^-1 (1) + 8 sin-1 (1)

= 16 sin^-1 (1)

= 16 × π/2

= 8π sq units

Calculation:

Enclosed Area

$=2{\int }_0^{\pi /4}\left(\cos x-\sin x\right)dx$

$=2{\left[\sin x+\cos x\right]}_0^{\pi /4}$

$=2\left[\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-\left(0+1\right)\right]$

$=2\left(\sqrt{2}-1\right)$

Concept:

Area under a Curve by Integration

Find the area under this curve by summing vertically.

• In this case, we find the area is the sum of the rectangles, height x = f(y) and width dy.
• If we are given y = f(x), then we need to re-express this as x = f(y) and we need to sum from the bottom to top.

So, $\mathbf{A}={\int }_{\mathbf{a}}^{\mathbf{b}}\mathbf{x}\mathbf{d}\mathbf{y}={\int }_{\mathbf{a}}^{\mathbf{b}}\mathbf{f}\left(\mathbf{y}\right)\mathbf{d}\mathbf{y}$

Calculation:

Given Curve: x = 4 - y2

⇒ y2 = 4 - x
⇒ y2 = - (x - 4)           

The above curve is the equation of the Parabola,

We know that at y-axis; x = 0

⇒ y2 = 4 - x

⇒ y2 = 4 - 0 = 4

⇒ y = ± 2

⇒ (0, 2) or (0, -2) are Point of intersection.

Area under the curve $={\int }_{-2}^2\mathrm{x}\mathrm{d}\mathrm{y}$

$={\int }_{-2}^2(4-{\mathrm{y}}^2)\mathrm{d}\mathrm{y}$

$={[4\mathrm{y}-\frac{{\mathrm{y}}^3}{3}]}_{-2}^2$

$=\frac{32}{3}$ Sq. unit

Explanation:

Equation of circle is given by x² + y² = a²

Let's take the strip along a y-direction and integrate it from 0 to 'a' this will give the area of the first quadrant and in order to find out the area of a circle multiply by 4

$y=\sqrt{x^2-a^2}$

Area of first Quadrant = $\underset{0}{\overset{a}{\int }}ydx$ = $\underset{0}{\overset{a}{\int }}\sqrt{a^2-x^2}dx$

Area of circle = 4 × $\underset{0}{\overset{a}{\int }}\sqrt{a^2-x^2}dx$

Given:

The parabolas y = 3x2 and x2 - y + 4 = 0

Concept:

Apply concept of area between two curves y₁ and y₂ between x = a and x = b

$\mathrm{A}={\int }_{\mathrm{a}}^{\mathrm{b}}({\mathrm{y}}_1-{\mathrm{y}}_2)\mathrm{d}\mathrm{x}$

Calculation:

The parabolas y = 3x2 and x2 - y + 4 = 0

then 3x² = x² + 4

⇒ x² = 2

⇒ x = ± √ 2

Then the area is 

$\mathrm{A}={\int }_{-\sqrt{2}}^{\sqrt{2}}({\mathrm{x}}^2+4-3{\mathrm{x}}^2)\mathrm{d}\mathrm{x}$

$\mathrm{A}={\int }_{-\sqrt{2}}^{\sqrt{2}}(4-2{\mathrm{x}}^2)\mathrm{d}\mathrm{x}$

$\mathrm{A}=[4\mathrm{x}-2\frac{{\mathrm{x}}^3}{3}{]}_{-\sqrt{2}}^{\sqrt{2}}$

$\mathrm{A}=4[\sqrt{2}-(-\sqrt{2})]-\frac{2}{3}[{\sqrt{2}}^3-{(-\sqrt{2})}^3]$

$\mathrm{A}=8\sqrt{2}-\frac{8}{3}\sqrt{2}$

$\mathrm{A}=\frac{16\sqrt{2}}{3}$ sq unit.

Hence option (4) is correct.

Concept:

The area of the curve y = f(x) is given by:

A = ${\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}$

where x1 and x2 are the endpoints between which the area is required.

Imp. Note: The net area will be the addition of the area below the x-axis and the area above the x-axis.

Calculation:

The f(x) = y = 4x³

Given the end points x1 = -2, x2 = 3

Area of the curve (A) = $\left|{\int }_{-2}^{3}4{\mathrm{x}}^3\mathrm{d}\mathrm{x}\right|$

⇒ A = $\left|{\int }_{-2}^{0}4{\mathrm{x}}^3\mathrm{d}\mathrm{x}\right|+\left|{\int }_0^{3}4{\mathrm{x}}^3\mathrm{d}\mathrm{x}\right|$

⇒ A = $\left|4{\left[\frac{{\mathrm{x}}^4}{4}\right]}_{-2}^0\right|+\left|4{\left[\frac{{\mathrm{x}}^4}{4}\right]}_0^3\right|$

⇒ A = $\left|[0-2^4]\right|+\left|[3^4-0]\right|$

⇒ A = $|-16|+\left|81\right|$

⇒ A = 97

Additional Information

Integral property:

• ∫ xn dx = $\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}$+ C ; n ≠ -1
• $\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\ln \mathrm{x}$ + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C 

Explanation:

Given equation of curves are

y = x²    ---(1) and y = 16    ---(2)

By solving both equation (1) and (2) we have:

x² = 16

x = 4, -4.

∴ Points of intersection are (4, 16) and (-4, 16).

From the figure we have,

$RequiredArea={\int }_{-4}^4(16-x^2)dx$

By using Integral property we have,

 $A=2{\int }_0^4(16-x^2)dx$

$=2{\left[16x-\frac{x^3}{3}\right]}_0^4$

$=2{\left[16x-\frac{x^3}{3}\right]}_0^4$

$=2\left[64-\frac{64}{3}\right]$

$=2\times 64\times \frac{2}{3}$

 $A=\frac{256}{3}sq.units$

Alternate Method 

There is another method also by which we can solved the problem,

By considering horizontal strip and by the condition of symmetry we have:

$Area=2{\int }_0^{16}xdy$

$Area=2{\int }_0^{16}\sqrt{y}dy$

$Area=2\times \frac{2}{3}\times [y^{\frac{3}{2}}{]}_0^{16}$

$Area=2\times \frac{2}{3}\times [{16}^{\frac{3}{2}}-0]$

Area = $\frac{256}{3}sq.unit$

Concept:

The area under a Curve by Integration:

Find the area under this curve is by summing horizontally.

In this case, we find the area is the sum of the rectangles, heights y = f(x) and width dx.

We need to sum from left to right.

∴ Area =  ${\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{y}\mathrm{d}\mathrm{x}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}$

Calculation: 

Here, we have to find the area of the region bounded by the curves y = x2, x-axis and ordinates x = -1 and x = 2

So, the area enclosed by the given curves is given by ${\int }_{-1}^2{\mathrm{x}}^2\mathrm{d}\mathrm{x}$

As we know that, $\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}$

Area = ${\int }_{-1}^2{\mathrm{x}}^2\mathrm{d}\mathrm{x}$

= ${\left[\frac{{\mathrm{x}}^3}{3}\right]}_{-1}^2$

= $[\frac{8}{3}-\frac{-1}{3}]=\frac{9}{3}=3$

Area = 3 sq. units.

Explanation:

Given curves are y = x - 1 and y2 = 2x + 6 

On solving, we get, 

y² = 2(y + 1) + 6

⇒ y² - 2y - 8 = 0

⇒ (y - 4)(y + 2) = 0

⇒ y = -2, 4

Now, we can find the area by

A = ${\int }_{-2}^4[y+1-(\frac{y^2}{2}-3)]dy$

= ${\int }_{-2}^4(4+y-\frac{y^2}{2})dy$

= ${[4y+\frac{y^2}{2}-\frac{y^3}{6}]}_{-2}^4$

= $16+8-\frac{32}{3}-(-8+2+\frac{4}{3})$

∴ A = 18