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:

\(\mathop \smallint \limits_0^1 \left| {5x - 3} \right|dx\)

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.

\(\mathop \smallint \limits_0^1 \left| {5x - 3} \right|dx = - \mathop \smallint \limits_0^{\frac{3}{5}} \left( {5x - 3} \right)dx + \mathop \smallint \limits_{\frac{3}{5}}^1 \left( {5x - 3} \right)dx\)

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

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

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

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:

\(\mathop \smallint \limits_0^1 \left| {5x - 3} \right|dx\)

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.

\(\mathop \smallint \limits_0^1 \left| {5x - 3} \right|dx = - \mathop \smallint \limits_0^{\frac{3}{5}} \left( {5x - 3} \right)dx + \mathop \smallint \limits_{\frac{3}{5}}^1 \left( {5x - 3} \right)dx\)

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

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

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

Let,

 \(I = \smallint {e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx\)

\( = \smallint {e^x}f\left( x \right)dx + \smallint {e^x}f'\left( x \right)dx+C\)

By solving through integration by parts, we get

\( = \left\{ {{e^x}f\left( x \right) - \smallint f'\left( x \right){e^x}dx} \right\} + \smallint {e^x}f'\left( x \right)dx +C\)

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

where C is constant

\(-{ x }^{ 2 }+3x-2=-{ x }^{ 2 }+2x+x-2=-x\left( x-2 \right) +1\left( x-2 \right) =\left( 1-x \right) \left( x-2 \right)\)

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 \(√ x/ √ x\) dx 

⇒ Put √x = t

⇒ \(\frac{1 }{2 √ x} dx \) = dt 

⇒ \(\frac{1 }{√ x} dx = 2dt\) 

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

Hence, The Correct Answer is Option 3.

Explanation:

\(\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\) is not true for n = -1

because for n = -1

\(\int x^{n} d x=\int x^{-1} d x\) = \(\int {1\over x}dx\) = 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 =  \(\frac{1}{4}\int {{(\sqrt {t+2})dt +\frac{1}{4}\int\sqrt tdt}}\)

⇒ 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,

\(\mathop \smallint \limits_0^1 \left[ {\left( {1 + f\left( x \right)} \right)x + \mathop \smallint \limits_0^x f\left( t \right)dt} \right]dx = 1\)     ----(1)

We know that,

\(\mathop \smallint \limits_0^x f\left( t \right)dt = {\left[ {\smallint f\left( x \right)dx} \right]_{x = x}} - {\left[ {\smallint f\left( x \right)dx} \right]_{x = 0}}\)

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

\(\mathop \smallint \limits_0^1 xf\left( x \right)dx = \left[ {x\smallint f\left( x \right)dx} \right]_0^1 - \mathop \smallint \limits_0^1 \left[ {\smallint f\left( x \right)dx} \right]dx\)      ----(3)

Now, equation (1) can be written as

\(\mathop \smallint \limits_0^1 x\;dx + \mathop \smallint \limits_0^1 xf\left( x \right)dx + \mathop \smallint \limits_0^1 \left( {\mathop \smallint \limits_0^x f\left( t \right)dt} \right)dx = 1\)

\( \Rightarrow \left[ {\frac{{{x^2}}}{2}} \right]_0^1 + \mathop \smallint \limits_0^1 x\;f\left( x \right)dx + \mathop \smallint \limits_0^1 \left( {\mathop \smallint \limits_0^x f\left( t \right)dt} \right)dx = 1\)

\( \Rightarrow \mathop \smallint \limits_0^1 xf\left( x \right)dx + \mathop \smallint \limits_0^1 \left( {\mathop \smallint \limits_0^x f\left( t \right)dt} \right)dx = \frac{1}{2}\)

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

\( \Rightarrow \left[ {x\smallint f\left( x \right)dx} \right]_0^1 - \mathop \smallint \limits_0^1 \left[ {\smallint f\left( x \right)dx} \right]dx + \mathop \smallint \limits_0^1 \left( {\smallint f\left( x \right)dx} \right)dx - \mathop \smallint \limits_0^1 {\left[ {\smallint f\left( x \right)dx} \right]_{x = 0}}dx = \frac{1}{2}\)

\( \Rightarrow \left[ {x\smallint f\left( x \right)dx} \right]_0^1 - \mathop \smallint \limits_0^1 {\left[ {\smallint f\left( x \right)dx} \right]_{x = 0}}dx = \frac{1}{2}\)

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

\( \Rightarrow \mathop \smallint \limits_0^1 f\left( x \right) = dx = \frac{1}{2}\)