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

Concept:

For Integration with modulus, first we have to find the point where the sign of the value of the function gets change.

Calculation:

Given:

$\underset{0}{\overset{1}{\int }}\left|5x-3\right|dx$

f(x) = 5x - 3 = 0

x = 3/5

∴ from 0 to 3/5 the function is negative and 3/5 to 1 the function is positive.

$\underset{0}{\overset{1}{\int }}\left|5x-3\right|dx=-\underset{0}{\overset{\frac{3}{5}}{\int }}\left(5x-3\right)dx+\underset{\frac{3}{5}}{\overset{1}{\int }}\left(5x-3\right)dx$

$={\left(-\frac{5}{2}{x}^2+3x\right)}_0^{\frac{3}{5}}+{\left(\frac{5x^2}{2}-3x\right)}_{\frac{3}{5}}^1$

$=\left(-\frac{9}{10}+\frac{9}{5}\right)+\left[\left(\frac{5}{2}-3\right)-\left(\frac{9}{10}-\frac{9}{5}\right)\right]$

$=\frac{9}{10}+\left(\frac{-1}{2}+\frac{9}{10}\right)=\frac{13}{10}$

Concept:

For Integration with modulus, first we have to find the point where the sign of the value of the function gets change.

Calculation:

Given:

$\underset{0}{\overset{1}{\int }}\left|5x-3\right|dx$

f(x) = 5x - 3 = 0

x = 3/5

∴ from 0 to 3/5 the function is negative and 3/5 to 1 the function is positive.

$\underset{0}{\overset{1}{\int }}\left|5x-3\right|dx=-\underset{0}{\overset{\frac{3}{5}}{\int }}\left(5x-3\right)dx+\underset{\frac{3}{5}}{\overset{1}{\int }}\left(5x-3\right)dx$

$={\left(-\frac{5}{2}{x}^2+3x\right)}_0^{\frac{3}{5}}+{\left(\frac{5x^2}{2}-3x\right)}_{\frac{3}{5}}^1$

$=\left(-\frac{9}{10}+\frac{9}{5}\right)+\left[\left(\frac{5}{2}-3\right)-\left(\frac{9}{10}-\frac{9}{5}\right)\right]$

$=\frac{9}{10}+\left(\frac{-1}{2}+\frac{9}{10}\right)=\frac{13}{10}$

Let,

 $I=\int e^x\left\{f\left(x\right)+f^{\prime }\left(x\right)\right\}dx$

$=\int e^{x}f\left(x\right)dx+\int e^x{f}^{\prime }\left(x\right)dx+C$

By solving through integration by parts, we get

$=\left\{e^{x}f\left(x\right)-\int f^{\prime }\left(x\right)e^{x}dx\right\}+\int e^x{f}^{\prime }\left(x\right)dx+C$

$=f\left(x\right).e^x+C$

where C is constant

Explanation:

To solve this, we can use

Let $u=\log x$  so that $du=\frac{1}{x}dx$

dv = dx , so that v = x

Now, using integration by parts $\int \log xdx=x\log x-\int x\cdot \frac{1}{x}dx$

$=x\log x-\int 1dx$

=$=x\log x-x+C$

= $x(\log x-1)+C$

So, the correct answer is $x(\log x-1)+C$

The correct option is option 3.

Explanation:

$\int \sqrt{1+\sin x}dx$

= $\int \sqrt{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}dx$

= $\int \sqrt{(\sin \frac{x}{2}+\cos \frac{x}{2}{)}^2}dx$

= $\int (\sin \frac{x}{2}+\cos \frac{x}{2})dx$

= 2(- cos (x/2) + sin (x/2))

= 2(sin (x/2) - cos (x/2))

(3) is true.

$\begin{array}{l}g\left(x\right)=\int f\left(x\right)dx\\ f\left(x\right)=e^{-x-e^{-x}}\\ g\left(x\right)=\int e^{-x-e^{-x}}dx\\ =\int \frac{e^{-x}}{e^{e^{-x}}}dx\end{array}$

Substitude e^-x = t

-e^-x dx = dt

$\begin{array}{l}g\left(x\right)=\int -\frac{dt}{e^t}\\ =\int -e^{-t}dt\\ g\left(x\right)=e^{-t}\\ g\left(x\right)=e^{-e^{-x}}\end{array}$

Explanation:

$I=\underset{0}{\overset{1}{\int }}\frac{{\left({\sin }^{-1}x\right)}^2}{\sqrt{1-x^2}}dx$

Put sin^-1 x = t

$\begin{array}{l}\frac{dx}{\sqrt{1-x^2}}=dt\\ =\underset{0}{\overset{\frac{\pi }{2}}{\int }}{t}^{2}dt\\ ={\frac{t^3}{3}]}_0^{\frac{\pi }{2}}\\ =\frac{{\pi }^3}{24}\end{array}$

Concept:

${\displaystyle {\int }_0^m{\int }_0^{x}f(x,y)dA}$

Since the inside limit is in terms of x, therefore we have to integrate first the 'y' terms and convert whole expression in terms of x.

Calculation:

Given:

${\displaystyle {\int }_0^1{\int }_0^x(x^2+y^2)dA}$

${\displaystyle {\int }_0^1[{\int }_0^x(x^2+y^2)dy]dx}$

${\displaystyle {\int }_0^1(x^2[y{]}_0^x+{\left[\frac{y^3}{3}\right]}_0^x)dx}$

${\displaystyle {\int }_0^1(x^3+\frac{x^3}{3})dx}$

${\displaystyle {\int }_0^1\left(\frac{4x^3}{3}\right)dx}$

${\displaystyle {\left(\frac{4x^4}{12}\right)}_0^1}$

${\displaystyle \frac{1}{3}}$

The Given Question is Wrong, Marks are allotted to all.

Explanation-

Given that, $I=\int (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+.....$

$I=\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+$......

Expanding the options given,

option 1-

$\frac{1}{1+x}=(1+x{)}^{-1}=1-x+x^2-x^3.......$

option 2-

$\frac{1}{1+x^2}=(1+x^2{)}^{-1}=1-x^2+x^4-x^6.......$

option 3-

$\frac{-1}{1-x}=-(1-x{)}^{-1}=-(1+x+x^2+x^3.......$

option 4-

$\frac{-1}{1-x^2}=-(1-x^2{)}^{-1}=-(1+x^2+x^4+x^6.......$

So none of the options are matching with the correct answer. 

Explanation:

Given Integral 

${\displaystyle \int {\mathrm{e}}^{\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\mathrm{x}}\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{d}\mathrm{x}}$

Let, sin^-1 x = t ⇒ $\frac{1}{\sqrt{1-x^2}}dx=dt$ as we know the derivative of sin-1 x  = $\frac{1}{\sqrt{1-x^2}}$ 

Now the equation reduces to 

${\displaystyle \int {\mathrm{e}}^{\mathrm{t}}\mathrm{d}\mathrm{t}}$ ⇒ we know ∫e^xdx = e^x + c 

∴ ∫ etdt = et + c, as t = sin-1 x our equation becomes e^sin-1x  + c

∴ ${\displaystyle \int {\mathrm{e}}^{\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\mathrm{x}}\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{d}\mathrm{x}}$ = ${\mathrm{e}}^{{\sin }^{-1}\mathrm{x}}+\mathrm{c}$

Given,

I = $\int \frac{(\cos x{)}^{1.5}-(\sin x{)}^{1.5}}{\sqrt{\sin x\cdot \cos x}}dx$

⇒ I = $\int \frac{\cos x}{\sqrt{\sin x}}-\frac{\sin x}{\sqrt{\cos x}}dx$

⇒ I = $\int \frac{\cos x}{\sqrt{\sin x}}dx-\int \frac{\sin x}{\sqrt{\cos x}}dx$ = I₁ - I₂ (let)

Computing I₁,

I₁ = $\int \frac{\cos x}{\sqrt{\sin x}}dx$

Put sin x = t → cos x dx = dt

⇒ I₁ = $\int \frac{dt}{\sqrt{t}}=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c_1$

⇒ I1 = $2\sqrt{t}+c_1$

⇒ I1 = $2\sqrt{\sin x}+c_1$

Similarly, 

I2 = $\int \frac{\sin x}{\sqrt{\cos x}}dx$

Explanation:

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$ is not true for n = -1

because for n = -1

$\int x^{n}dx=\int x^{-1}dx$ = $\int \frac{1}{x}dx$ = log |x| + c

(5) is correct

Explanation:

$I=\underset{0}{\overset{\mathrm{\infty }}{\int }}x{e}^{-x^2}dx$

Let, t = x² ⇒ dt = 2xdx ⇒ xdx = dt/2

$\begin{array}{l}I=\frac{1}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-t}dt=\frac{1}{2}{\left|-e^{-t}\right|}_0^{\mathrm{\infty }}=-\frac{1}{2}{\left|\frac{1}{e^t}\right|}_0^{\mathrm{\infty }}=-\frac{1}{2}\left[0-1\right]\\ I=\frac{1}{2}\end{array}$  

Concept:

For Integration with modulus, first we have to find the point where the sign of the value of the function gets change.

Calculation:

Given:

$\underset{0}{\overset{1}{\int }}\left|5x-3\right|dx$

f(x) = 5x - 3 = 0

x = 3/5

∴ from 0 to 3/5 the function is negative and 3/5 to 1 the function is positive.

$\underset{0}{\overset{1}{\int }}\left|5x-3\right|dx=-\underset{0}{\overset{\frac{3}{5}}{\int }}\left(5x-3\right)dx+\underset{\frac{3}{5}}{\overset{1}{\int }}\left(5x-3\right)dx$

$={\left(-\frac{5}{2}{x}^2+3x\right)}_0^{\frac{3}{5}}+{\left(\frac{5x^2}{2}-3x\right)}_{\frac{3}{5}}^1$

$=\left(-\frac{9}{10}+\frac{9}{5}\right)+\left[\left(\frac{5}{2}-3\right)-\left(\frac{9}{10}-\frac{9}{5}\right)\right]$

$=\frac{9}{10}+\left(\frac{-1}{2}+\frac{9}{10}\right)=\frac{13}{10}$

$\begin{array}{l}g\left(x\right)=\int f\left(x\right)dx\\ f\left(x\right)=e^{-x-e^{-x}}\\ g\left(x\right)=\int e^{-x-e^{-x}}dx\\ =\int \frac{e^{-x}}{e^{e^{-x}}}dx\end{array}$

Substitude e^-x = t

-e^-x dx = dt

$\begin{array}{l}g\left(x\right)=\int -\frac{dt}{e^t}\\ =\int -e^{-t}dt\\ g\left(x\right)=e^{-t}\\ g\left(x\right)=e^{-e^{-x}}\end{array}$