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

Given:

$I={\int }_1^{2}ydx$

Since the integration is done w.r.t 'x' variable, 'y' in terms of x can be written as:

y = mx + c

The slope of the given line will be:

$m=\frac{1-0}{0-(-1)}=1$

y = x + c

when x = 0, y = 1, i.e.

c = 1

∴ y = x + 1

$I={\int }_1^{2}ydx={\int }_1^2\left(x+1\right)dx$

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

= 2.5

Concept:

${\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx={\int }_0^{\frac{\pi }{2}}{\cos }^{n}xdx$

If n is even, $\frac{\left(n-1\right)\left(n-3\right)\left(n-5\right)\dots \dots }{n\left(n-2\right)\left(n-4\right)\left(n-6\right)\dots ..}\times \frac{\pi }{2}$

If n is odd, $\frac{\left(n-1\right)\left(n-3\right)\left(n-5\right)}{n\left(n-2\right)\left(n-4\right)\left(n-6\right)}\times 1$                                            
Here n = 7, means n is odd,

Therefore, ${\int }_0^{\frac{\pi }{2}}{\sin }^{9}xdx=\frac{\left(9-1\right)\left(9-3\right)(9-5)(9-7)}{9\left(9-2\right)\left(9-4\right)\left(9-6\right)(9-8)}\times 1=0.4063$

Concept:

Use integration by parts for solving this problem.

Calculation:

${\int }_1^e\left(logxx\right)dx$

$={\{\log x.\frac{x^2}{2}\}}_1^e-{\int }_1^e\frac{1}{x}.\frac{x^2}{2}dx$ 

$={\left\{\frac{x^2\log x}{2}\right\}}_1^e-\frac{1}{2}{\int }_1^{e}xdx$ 

$={\left\{\frac{x^2\log x}{2}\right\}}_1^e-{\left\{\frac{x^2}{4}\right\}}_1^e$ 

$=\{\frac{e^2\log e}{2}-\frac{\log 1}{2}\}-\{\frac{e^2}{4}-\frac{1}{4}\}$ 

$=\frac{e^2}{2}-\frac{e^2}{4}+\frac{1}{4}$ 

$\Rightarrow \frac{e^2+1}{4}=2.097$

Concept:

Some important relations

1) ${\mathrm{e}}^{\mathrm{i}\mathrm{x}}=\cos \mathrm{x}+\mathrm{i}\sin \mathrm{x}$

2) ${\mathrm{e}}^{-\mathrm{i}\mathrm{x}}=\cos \mathrm{x}-\mathrm{i}\sin \mathrm{x}$

3) ${\mathrm{e}}^{\mathrm{i}\pi }=\cos \pi +\mathrm{i}\sin \pi =-1$

4) $\frac{-1}{\mathrm{i}}=\mathrm{i}$

Calculation:

$\begin{array}{l}I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}+\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}-\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}\mathrm{d}\mathrm{x}\\ =\underset{0}{\overset{\frac{\pi }{0}}{\int }}\frac{e^{ix}}{e^{-ix}}dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{e}^{2ix}dx\\ =\frac{1}{2i}{\left[e^{2ix}\right]}_0^{\frac{\pi }{2}}=\frac{1}{2i}\left[e^{i\pi }-1\right]\\ =\frac{1}{2i}\left(-1-1\right)\\ =-\frac{1}{i}\\ =i\end{array}$

Concept:

If f(x)  is odd function then f(-x) = -f(x)

Properties of definite integral

If f(x)  is even function then ${\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=2{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}$

If f(x)  is odd function then ${\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=0$

Calculation:

Let I = ${\displaystyle {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\displaystyle \frac{\mathrm{s}\mathrm{i}{\mathrm{n}}^3\mathrm{x}{\mathrm{e}}^{-{\mathrm{x}}^2}}{{\mathrm{x}}^4}}\mathrm{d}\mathrm{x}}$

Let f(x) = ${\displaystyle \frac{si{n}^{3}x{e}^{-x^2}}{x^4}}$

Replaced x by -x, 

⇒ f(-x) = ${\displaystyle \frac{\mathrm{s}\mathrm{i}{\mathrm{n}}^3(-\mathrm{x}){\mathrm{e}}^{-(-\mathrm{x}{)}^2}}{(-\mathrm{x}{)}^4}}$

As we know sin (-θ) = - sin θ

= ${\displaystyle \frac{-si{n}^{3}x{e}^{-x^2}}{x^4}}$

⇒ f(-x) = -f(x)      

So, f(x) is odd function

Therefore, I = 0     

Explanation:

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

$\mathrm{L}\mathrm{e}\mathrm{t}f\left(x\right)=x{e}^{-x^2}$

$f\left(-x\right)=-x{e}^{-x^2}=-f\left(x\right)$

∴ f(x) is odd function

∴ The value of given integral is zero.

Concept:

Improper Integral: If a function f on [a, b] have infinite value then it is called is improper integral

1. Improper Integral of the First kind: ${\displaystyle {\int }_a^{b}f\left(x\right)dx}$ is said to be the improper integral of the first kind if a = -∞ or b = ∞ or both.
2. Improper Integral of the second kind: ${\displaystyle {\int }_a^{b}f\left(x\right)dx}$ is said to be the improper integral of the second kind if a or b is finite but f(x) is infinite for some x ∈ [a, b].

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If the integration of the improper integral exists, then it is called as Converges, But if the limit of integration fails to exist, then the improper integral is said to be Diverge.

${\displaystyle {\int }_a^b\frac{dx}{(b-x{)}^p}=\{\begin{array}{c}\mathrm{\infty };p\ge 1\\ \frac{1}{(1-p)(b-a{)}^{(p-1)}};p<1\end{array}}$          ---(1)

For this type, if p ≥ 1 the function is divergent and for p < 1, the function is convergent.

${\displaystyle {\int }_a^b\frac{dx}{(x-a{)}^p}=\{\begin{array}{c}\mathrm{\infty };p\ge 1\\ \frac{1}{(1-p)(b-a{)}^{(p-1)}};p<1\end{array}}$       ---(2)

For this type, if p ≥ 1 the function is divergent and for p < 1, the function is convergent.

Calculation:

Given:

${\displaystyle \underset{2}{\overset{5}{\int }}\frac{dx}{{(x-2)}^{\frac{1}{2}}}}$

On comparing with equation (2) the value of p < 1, therefore it is convergent.

${\displaystyle \underset{2}{\overset{5}{\int }}\frac{dx}{{(x-2)}^{\frac{1}{2}}}=\frac{1}{((1-\frac{1}{2})(5-2{)}^{(\frac{1}{2}-1)})}}$

$=\frac{1}{\frac{1}{2}\times 3^{-\frac{1}{2}}}=2\sqrt{3}$

Analysis:

$\int {\sin }^5\left(x\right)\cdot {\cos }^2\left(x\right)dx$

This can be written as:

$=\int \sin \left(x\right)\cdot {\left({\sin }^2\left(x\right)\right)}^2\cdot {\cos }^2\left(x\right)dx$  ---(1)

Since sin²x + cos²x = 1:

sin2x = 1 - cos2x

Equation (1) can now be written as:

$\int \mathrm{s}\mathrm{i}\mathrm{n}\left(x\right)\cdot {\left(1-{\cos }^2\left(x\right)\right)}^2\cdot {\cos }^2\left(x\right)dx$

Using trigonometric substitution, let u = cos (x)

$\frac{du}{dx}=-\sin \left(x\right)$

$dx=\frac{du}{-\sin \left(x\right)}$

$\int \frac{\sin \left(x\right)\cdot {\left(1-u^2\right)}^2\cdot u^2}{-\sin \left(x\right)}du=-\int u^2\cdot {\left(1-u^2\right)}^{2}du$

$=-\int u^2\cdot \left(1+u^4-2u^2\right)du$

$=-\int (u^2+u^6-2u^4)du$

= $-\int u^{2}du-\int u^{6}du-\int -2u^{4}du$

$=-\frac{1}{3}{u}^3-\frac{1}{7}{u}^7+\frac{2}{5}{u}^5+C$

Substituting u = cos x back in the above expression, we get:

$=\frac{2}{5}{\cos }^5\left(x\right)-\frac{1}{3}{\cos }^3\left(x\right)-\frac{1}{7}{\cos }^7\left(x\right)+C$

Concept:

We can solve this by using integration by parts,

∫u dv = uv - ∫ v du

let u = x; dv/dx = cos x

⇒ $\underset{0}{\overset{\pi /2}{\int }}x\cdot \cos x\cdot dx=\int udv$

⇒ ${\left[x\cdot \left(\sin x\right)\right]}_0^{\pi /2}-\underset{0}{\overset{\pi /2}{\int }}\sin x\cdot dx$

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

$\Rightarrow \left(\frac{\pi }{2}\times 1+0\right)-\left(0+1\right)$

$\Rightarrow \frac{\pi }{2}-1$

Concept:

Integration: 

$\int \frac{1}{{\mathrm{a}}^2+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\frac{1}{\mathrm{a}}{\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}$.

Calculation:

The given expression is $\int \frac{1}{4+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\int \frac{1}{2^2+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}$.

Using $\int \frac{1}{{\mathrm{a}}^2+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\frac{1}{\mathrm{a}}{\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}$, we get:

$\int \frac{1}{4+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\int \frac{1}{2^2+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\frac{1}{2}{\tan }^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}$.