Integral Calculus MCQ Quiz - Objective Question with Answer for Integral Calculus - Download Free PDF

Calculation:

$\int x^2{e}^{-2x}dx=e^{-2x}(a{x}^2+bx+c)+D$

On differentiating both sides, we get

$x^2{e}^{-2x}=e^{-2x}(2ax+b)+(a{x}^2+bx+c)(-2e^{-2x})$

$=e^{-2x}[-2a{x}^2+2(a-b)]+(x+b-2c)$

$\Rightarrow a=1,2(a-b)=0,b-2c=0$

$\Rightarrow a=1,b=1\text{ and }c=\frac{1}{2}$

Hence, the correct answer is Option 4.

Calculation:

$\int x^2{e}^{-2x}dx=e^{-2x}(a{x}^2+bx+c)+D$

On differentiating both sides, we get

$x^2{e}^{-2x}=e^{-2x}(2ax+b)+(a{x}^2+bx+c)(-2e^{-2x})$

$=e^{-2x}[-2a{x}^2+2(a-b)]+(x+b-2c)$

$\Rightarrow a=1,2(a-b)=0,b-2c=0$

$\Rightarrow a=1,b=1\text{ and }c=\frac{1}{2}$

Hence, the correct answer is Option 3.

Calculation:

$\int x^2{e}^{-2x}dx=e^{-2x}(a{x}^2+bx+c)+D$

On differentiating both sides, we get

$x^2{e}^{-2x}=e^{-2x}(2ax+b)+(a{x}^2+bx+c)(-2e^{-2x})$

$=e^{-2x}[-2a{x}^2+2(a-b)]+(x+b-2c)$

$\Rightarrow a=1,2(a-b)=0,b-2c=0$

$\Rightarrow a=1,b=1\text{ and }c=\frac{1}{2}$

Hence, the correct answer is Option 1.

Calculation

Given $f(0)=0$ & $f^{\prime }(1)=0$ & $f(1)=\frac{1}{2}$

$\therefore f(x)=\frac{2x-x^2}{2}$

$f^{-1}(x)$ is the image of $f(x)$ on $y=x.$

Also, $2x+2y=3$ passes through $A(1,\frac{1}{2})$ & $B(\frac{1}{2},1)$

so bounded Area $A$

 $=AreaOAB=2[AreaOCM+AreaCMNA-AreaONA]$

$A=2[\frac{1}{2}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{2}(\frac{3}{4}+\frac{1}{2})\times \frac{1}{4}-\frac{1}{2}{\int }_0^1(2x-x^2),dx]$

$\Rightarrow A=2[\frac{9}{32}+\frac{5}{32}-[x^2-\frac{x^3}{3}{]}_0^1]$

$\Rightarrow A=2[\frac{14}{32}-(1-\frac{1}{3})]$
$\Rightarrow A=\frac{28}{32}-\frac{2}{3}=\frac{7}{8}-\frac{2}{3}$$\Rightarrow A=\frac{21-16}{24}=\frac{5}{24}$

⇒ 48A = 10

Calculation:

Given,

The function is $f(x)=\lfloor x^2\rfloor$.

We are tasked with finding the value of ${\int }_{\sqrt{2}}^{2}f(x)dx$.

We can break the integral into two parts as follows:

${\int }_{\sqrt{2}}^{2}f(x)dx={\int }_{\sqrt{2}}^{\sqrt{3}}2dx+{\int }_{\sqrt{3}}^{2}3dx$

For the range $\sqrt{2}\le x\le \sqrt{3}$, $\lfloor x^2\rfloor =2$. So the first part of the integral is:

${\int }_{\sqrt{2}}^{\sqrt{3}}2dx=2\times (\sqrt{3}-\sqrt{2})$

For the range $\sqrt{3}\le x\le 2$, $\lfloor x^2\rfloor =3$. So the second part of the integral is:

${\int }_{\sqrt{3}}^{2}3dx=3\times (2-\sqrt{3})$

Now, we calculate the values:

$2\times (\sqrt{3}-\sqrt{2})=2\sqrt{3}-2\sqrt{2}$

$3\times (2-\sqrt{3})=6-3\sqrt{3}$

Combining the two parts:

$2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}=6-2\sqrt{2}-\sqrt{3}$

Hence, the correct answer is Option 1.

Concept:

Definite Integral properties:

$\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{a}+\mathrm{b}-\mathrm{x}\right)\mathrm{d}\mathrm{x}$
Calculation:

Let f(x) = x(1 – x)⁹

Now using property, $\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{a}+\mathrm{b}-\mathrm{x}\right)\mathrm{d}\mathrm{x}$

${\int }_0^1\mathrm{x}(1-\mathrm{x}{)}^9\mathrm{d}\mathrm{x}=\underset{0}{\overset{1}{\int }}\left(1-\mathrm{x}\right){\left\{1-\left(1-\mathrm{x}\right)\right\}}^9\mathrm{d}\mathrm{x}$

$\Rightarrow \underset{0}{\overset{1}{\int }}\left(1-\mathrm{x}\right){\mathrm{x}}^9\mathrm{d}\mathrm{x}$

$\Rightarrow \underset{0}{\overset{1}{\int }}\left({\mathrm{x}}^9-{\mathrm{x}}^{10}\right)\mathrm{d}\mathrm{x}$

$\Rightarrow {\left[\frac{{\mathrm{x}}^{10}}{10}-\frac{{\mathrm{x}}^{11}}{11}\right]}_0^1$

⇒ 1/10 – 1/11

⇒ 1/110

∴ The value of integral ${\int }_0^1\mathrm{x}(1-\mathrm{x}{)}^9\mathrm{d}\mathrm{x}$ is 1/110.

Concept:

$\int \frac{\mathrm{d}\mathrm{x}}{{\mathrm{x}}^2+{\mathrm{a}}^2}=\frac{1}{\mathrm{a}}{\tan }^{-1}(\frac{\mathrm{x}}{\mathrm{a}})+\mathrm{c}$

Calculation:

Let I = ${\displaystyle {\int }_0^2{\displaystyle \frac{\mathrm{d}\mathrm{x}}{{\mathrm{x}}^2+4}}}$

= ${\displaystyle {\int }_0^2{\displaystyle \frac{\mathrm{d}\mathrm{x}}{{\mathrm{x}}^2+2^2}}}$

$=[\frac{1}{2}{\tan }^{-1}(\frac{\mathrm{x}}{2}){]}_0^2$

$=[\frac{1}{2}{\tan }^{-1}1-\frac{1}{2}{\tan }^{-1}0]$

$=\frac{1}{2}\times \frac{\pi }{4}-0$

= ${\displaystyle \frac{\pi }{8}}$

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:

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

Calculation: 

I = ${\int }_0^1\mathrm{x}\sqrt{{\mathrm{x}}^2+4}\mathrm{d}\mathrm{x}$

Let x² + 4 = t

Differentiating with respect to x, we get

⇒ 2xdx = dt

⇒ xdx = $\frac{\mathrm{d}\mathrm{t}}{2}$

Now,

I = $\frac{1}{2}{\int }_4^5\sqrt{\mathrm{t}}\mathrm{d}\mathrm{t}$

= $\frac{1}{2}{\left[\frac{{\mathrm{t}}^{3/2}}{\frac{3}{2}}\right]}_4^5$

= $\frac{1}{3}[5^{3/2}-4^{3/2}]$

= $\frac{1}{3}[5\sqrt{5}-8]$

Concept:

1 + cos 2x = 2cos² x

1 - cos 2x = 2sin² x

$\int \cos \mathrm{x}\mathrm{d}\mathrm{x}=\sin \mathrm{x}+\mathrm{c}$

Calculation:

I = $\int \mathrm{c}\mathrm{o}{\mathrm{s}}^2\mathrm{x}\mathrm{d}\mathrm{x}$

= $\int \frac{1+\cos 2\mathrm{x}}{2}\mathrm{d}\mathrm{x}$

= $\frac{1}{2}\int (1+\cos 2\mathrm{x})\mathrm{d}\mathrm{x}$

= $\frac{1}{2}[\mathrm{x}+\frac{\sin 2\mathrm{x}}{2}]+\mathrm{c}$

= $\frac{\mathrm{x}}{2}+\frac{\sin 2\mathrm{x}}{4}+\mathrm{c}$

Concept:

${\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{a}+\mathrm{b}-\mathrm{x}\right)\mathrm{d}\mathrm{x}$

Calculation: 

Let I = ${\int }_0^{2\pi }\frac{\sin 2\mathrm{x}}{\mathrm{a}-\mathrm{b}\cos \mathrm{x}}\mathrm{d}\mathrm{x}$         ----(1)

Using property f(a + b – x),

I = ${\int }_0^{2\pi }\frac{\sin 2(2\pi -\mathrm{x})}{\mathrm{a}-\mathrm{b}\cos (2\pi -\mathrm{x})}\mathrm{d}\mathrm{x}$   

As we know,  sin (2π - x) = - sin x and cos (2π - x) = cos x

I = ${\int }_0^{2\pi }\frac{-\sin 2\mathrm{x}}{\mathrm{a}-\mathrm{b}\cos \mathrm{x}}\mathrm{d}\mathrm{x}$         ----(2)       

I = -I

2I = 0

∴ I = 0

Concept:

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

Calculation: 

I = ${\int }_1^{\mathrm{\infty }}\frac{4}{{\mathrm{x}}^4}\mathrm{d}\mathrm{x}$

= ${\int }_1^{\mathrm{\infty }}4{\mathrm{x}}^{-4}\mathrm{d}\mathrm{x}$

= ${\left[\frac{4{\mathrm{x}}^{-3}}{-3}\right]}_1^{\mathrm{\infty }}$

= $\frac{-4}{3}{\left[\frac{1}{{\mathrm{x}}^3}\right]}_1^{\mathrm{\infty }}$

= $\frac{-4}{3}[\frac{1}{\mathrm{\infty }}-\frac{1}{1}]$

= $\frac{-4}{3}[0-1]$

= $\frac{4}{3}$

Concept:

${\displaystyle {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}={\displaystyle {\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})\mathrm{d}\mathrm{x}}}$

Calculations:

Consider, I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{\sin \mathrm{x}}}{\sqrt{\sin \mathrm{x}}+\sqrt{\cos \mathrm{x}}}}\mathrm{d}\mathrm{x}}$             ....(1)

I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{\sin ({\displaystyle \frac{\pi }{2}}-\mathrm{x})}}{\sqrt{\sin ({\displaystyle \frac{\pi }{2}}-\mathrm{x})}+\sqrt{\cos ({\displaystyle \frac{\pi }{2}}-\mathrm{x})}}}\mathrm{d}\mathrm{x}}$

I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{\cos \mathrm{x}}}{\sqrt{\cos \mathrm{x}}+\sqrt{\sin \mathrm{x}}}}\mathrm{d}\mathrm{x}}$                           ....(2)

Adding (1) and (2), we have

2I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{\cos \mathrm{x}}+\sqrt{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}}{\sqrt{\cos \mathrm{x}}+\sqrt{\sin \mathrm{x}}}}\mathrm{d}\mathrm{x}}$

2I = ${\displaystyle {\int }_0^{\pi /2}\mathrm{d}\mathrm{x}}$

2I = $[\mathrm{x}{]}_0^{\frac{\pi }{2}}$

I = ${\displaystyle \frac{\pi }{4}}$

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

Concept:

$\int \frac{1}{\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}\mathrm{d}\mathrm{x}={\sin }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$

Calculation:

I = $\int \frac{1}{\sqrt{16-25{\mathrm{x}}^2}}\mathrm{d}\mathrm{x}$

= $\int \frac{1}{\sqrt{16-(5\mathrm{x}{)}^2}}\mathrm{d}\mathrm{x}$

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = $\frac{\mathrm{d}\mathrm{t}}{5}$

Now,

I = $\frac{1}{5}\int \frac{1}{\sqrt{4^2-{\mathrm{t}}^2}}\mathrm{d}\mathrm{t}$

= $\frac{1}{5}{\sin }^{-1}\left(\frac{\mathrm{t}}{4}\right)$ + c

= $\frac{1}{5}{\sin }^{-1}\left(\frac{5\mathrm{x}}{4}\right)$ + c