Indefinite Integrals MCQ Quiz in मल्याळम - Objective Question with Answer for Indefinite Integrals - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is option '1'.

Concept:

Indefinite integral:

• An indefinite integral is the integration of a function without limits.
• Integration is the reverse process of differentiation.
• Integration is defined for a function f(x) and it helps in finding the area enclosed by the curve, with the reference to one of the coordinate axes.

Calculation:

Integration of e^kx 

Integrating f (x) = ekx with respect to dx.

$\int f(x)dx=\int e^{kx}dx$

$=\frac{e^{kx}}{k}+c$

where 'c' is the constant,

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

$-x^2+3x-2=-x^2+2x+x-2=-x(x-2)+1(x-2)=(1-x)(x-2)$

has real roots $1$ and $2$

So from case III: $\sqrt{-x^2+3x-2}=u(1-x)$ or $u(x-2)$

Given
We need to evaluate the integral ∫sin(√x/√/x)dx

Formula Used
using the Substitution method for integration.

Calculation:
∫ sin $\surd x/\surd x$ dx 

⇒ Put √x = t

⇒ $\frac{1}{2\surd x}dx$ = dt 

⇒ $\frac{1}{\surd x}dx=2dt$ 

$\int$ sin $\surd x/\surd x$ dx = 2 ∫ sin t dt = -2 cost + c = -2 cos(√x) + c

Hence, The Correct Answer is Option 3.

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:

Let I = $\int \frac{dx}{\sqrt{2x+5}-\sqrt{2x+3}}$

 ⇒ I = $\int \frac{dx}{\sqrt{2x+3+2}-\sqrt{2x+3}}$

Let 2x + 3 = t ⇒ 2 dx = dt  

dx = $\frac{dt}{2}$

⇒ I = $\int \frac{dt}{2(\sqrt{t+2}-\sqrt{t})}$

⇒ l=  $\frac{1}{2}\int \frac{dt}{\sqrt{t+2}-\sqrt{t}}$

⇒ I = $\frac{1}{2}\int \frac{(\sqrt{t+2}+\sqrt{t})dt}{(\sqrt{t+2}-\sqrt{t})(\sqrt{t+2}+\sqrt{t})}$

⇒ I = $\frac{1}{2}\int \frac{(\sqrt{t+2}+\sqrt{t})dt}{t+2-t}$

⇒ I = $\frac{1}{2}\int \frac{(\sqrt{t+2}+\sqrt{t})dt}{2}$

⇒ I = $\frac{1}{4}\int (\sqrt{t+2}+\sqrt{t})dt$

⇒ I =  $\frac{1}{4}\int (\sqrt{t+2})dt+\frac{1}{4}\int \sqrt{t}dt$

⇒ I = $\frac{1}{4}[\frac{(t+2{)}^{\frac{1}{2}+1}}{(\frac{1}{2}+1)}]$ + $\frac{1}{4}[\frac{(t{)}^{\frac{1}{2}+1}}{(\frac{1}{2}+1)}]$ +C

⇒ I = $\frac{1}{4}[\frac{(t+2{)}^{\frac{3}{2}}}{\frac{3}{2}}]$ + $\frac{1}{4}[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}]$ + C

⇒ I = $\frac{1}{4}\times \frac{2}{3}$ $[(t+2{)}^{\frac{3}{2}}]$ + $\frac{1}{4}\times \frac{2}{3}$$[t^{\frac{3}{2}}]$ +C

⇒ I = $\frac{1}{6}$ $[(t+2{)}^{\frac{3}{2}}]$ + $\frac{1}{6}$ $[t^{\frac{3}{2}}]$ +C

⇒ I=  $\frac{1}{6}$ [$(t+2{)}^{\frac{3}{2}}+t^{\frac{3}{2}}$ ]+C  

Substitute t = 2x + 3 in the above equation, we have

⇒ I = $\frac{1}{6}$ [$(2x+3+2{)}^{\frac{3}{2}}+(2x+3{)}^{\frac{3}{2}}$ ] +C

⇒ I =  $\frac{1}{6}$[ $(2x+5{)}^{\frac{3}{2}}+(2x+3{)}^{\frac{3}{2}}$ ] +C

⇒ $\int \frac{dx}{\sqrt{2x+5}-\sqrt{2x+3}}$ = $\frac{1}{6}$ [$(2x+5{)}^{\frac{3}{2}}+(2x+3{)}^{\frac{3}{2}}$] +C

Hence, the correct answer is option (3).

Given that,

$\underset{0}{\overset{1}{\int }}\left[\left(1+f\left(x\right)\right)x+\underset{0}{\overset{x}{\int }}f\left(t\right)dt\right]dx=1$     ----(1)

We know that,

$\underset{0}{\overset{x}{\int }}f\left(t\right)dt={\left[\int f\left(x\right)dx\right]}_{x=x}-{\left[\int f\left(x\right)dx\right]}_{x=0}$

$=\int f\left(x\right)dx-{\left[\int f\left(x\right)dx\right]}_{x=0}$      ----(2)

$\underset{0}{\overset{1}{\int }}xf\left(x\right)dx={\left[x\int f\left(x\right)dx\right]}_0^1-\underset{0}{\overset{1}{\int }}\left[\int f\left(x\right)dx\right]dx$      ----(3)

Now, equation (1) can be written as

$\underset{0}{\overset{1}{\int }}xdx+\underset{0}{\overset{1}{\int }}xf\left(x\right)dx+\underset{0}{\overset{1}{\int }}\left(\underset{0}{\overset{x}{\int }}f\left(t\right)dt\right)dx=1$

$\Rightarrow {\left[\frac{x^2}{2}\right]}_0^1+\underset{0}{\overset{1}{\int }}xf\left(x\right)dx+\underset{0}{\overset{1}{\int }}\left(\underset{0}{\overset{x}{\int }}f\left(t\right)dt\right)dx=1$

$\Rightarrow \underset{0}{\overset{1}{\int }}xf\left(x\right)dx+\underset{0}{\overset{1}{\int }}\left(\underset{0}{\overset{x}{\int }}f\left(t\right)dt\right)dx=\frac{1}{2}$

By substituting equation (2) and (3) in the above equation

$\Rightarrow {\left[x\int f\left(x\right)dx\right]}_0^1-\underset{0}{\overset{1}{\int }}\left[\int f\left(x\right)dx\right]dx+\underset{0}{\overset{1}{\int }}\left(\int f\left(x\right)dx\right)dx-\underset{0}{\overset{1}{\int }}{\left[\int f\left(x\right)dx\right]}_{x=0}dx=\frac{1}{2}$

$\Rightarrow {\left[x\int f\left(x\right)dx\right]}_0^1-\underset{0}{\overset{1}{\int }}{\left[\int f\left(x\right)dx\right]}_{x=0}dx=\frac{1}{2}$

$\Rightarrow {\left[\int f\left(x\right)dx\right]}_{x=1}-{\left[\int f\left(x\right)dx\right]}_{x=0}=\frac{1}{2}$

$\Rightarrow \underset{0}{\overset{1}{\int }}f\left(x\right)=dx=\frac{1}{2}$