2 Marks Questions

Question 1512 Marks

Evaluate the following integrals:
$\int\limits^1_0\text{xe}^{\text{x}^2}\text{dx}$

Answer

We have,
$\text{I}=\int\limits^1_0\text{xe}^{\text{x}^2}\text{dx}=\frac{1}{2}\int\limits^1_0\text{e}^{\text{x}^2}2\text{dx}$
Putting $\text{x}^2=\text{z}$
$2\text{x dx}=\text{dz}$
When $\text{x}\rightarrow0;\text{ z}\rightarrow0$
And $\text{x}\rightarrow1;\text{ z}\rightarrow1$
$\therefore\ \text{I}=\frac{1}{2}\int\limits^1_0\text{e}^2\text{ dz}$
$=\frac{1}{2}\times\big[\text{e}^{\text{z}}\big]^1_0$
$=\frac{1}{2}(\text{e}-\text{e}^0)$
$=\frac{1}{2}(\text{e}-1)$

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Question 1522 Marks

Evaluate the following integrals:
$\int\frac{1}{3\sqrt{\text{x}^2}}\text{dx}$

Answer

$\int\frac{\text{dx}}{3\sqrt{\text{x}^2}}$
$=\int\frac{\text{dx}}{\text{x}^\frac{2}{3}}$
$=\int\text{x}^\frac{-2}{3}\text{dx}$
$=\frac{\text{x}^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+\text{c}$
$=3\text{x}^\frac{1}{3}+\text{c}$

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Question 1532 Marks

Evaluate the following integrals:
$\int\big(\text{x}^\text{e}+\text{e}^\text{x}+\text{e}^\text{e}\big)\text{dx}$

Answer

$\int\big(\text{x}^\text{e}+\text{e}^\text{x}+\text{e}^\text{e}\big)\text{dx}$
$=\int\text{x}^\text{e}\text{dx}+\int\text{e}^\text{x}\text{dx}+\text{e}^\text{e}\int1\text{dx}$
$=\frac{\text{x}^{\text{e}+1}}{\text{e}+1}+\text{e}^\text{x}+\text{x}.\text{e}^\text{e}+\text{C}$

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Question 1542 Marks

Evaluate $\int(1-\text{x})\sqrt{\text{x}}\text{ dx}$

Answer

Let $\sqrt{\text{x}}=\text{t}$
$\text{x}=\text{t}^2$
$1-\text{x}=1-\text{t}^2$
$-\text{dx}=-2\text{tdt}$
$\text{dx}=2\text{tdt}$
$\text{I}=\int(1-\text{x})\sqrt{\text{x}}\text{ dx}$
$=\int(1-\text{t}^2)\text{t}2\text{t dt}$
$=2\int (1-\text{t}^2)\text{t}^2\text{ dt}$
$=2\big(\int\text{t}^2\text{dt}-\int\text{t}^4\text{dt}\big)$
$=2\frac{\text{t}^3}{3}-2\frac{\text{t}^5}{5}+\text{C}$
$=\frac{2}{3}\text{x}^{\frac{3}{2}}-\frac{2}{5}\text{x}^{\frac{5}{2}}+\text{C}$

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Question 1552 Marks

Write tha value of $\int\sec\text{x}(\sec\text{x}+\tan\text{x})\text{dx}$

Answer

$\int\sec\text{x}(\sec\text{x}+\tan\text{x})\text{dx}$
$=\int(\sec^2\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
$=\tan\text{x}+\sec\text{x}+\text{C}$

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Question 1562 Marks

Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$

Answer

Let $\text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
Here, $\text{f}(\theta)=\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)$
Consider, $\text{f}(-\theta)=\log\bigg[\frac{\text{a}-\sin(-\theta)}{\text{a}+\sin(-\theta)}\bigg]$
$=-\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)=-\text{f}(\theta)$
i.e., $\text{f}(\theta)$ is odd function.
Therefore, $\text{I}=0$

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Question 1572 Marks

Evaluate the following integrals:
$\int\limits^1_0\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$

Answer

We have,
$\text{I}=\int\limits^1_0\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
Putting $1+\text{x}^2=\text{t}$
$2\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0;\text{ t}\rightarrow1$
And $\text{x}\rightarrow1;\text{ t}\rightarrow2$
$\therefore\ \text{I}=\int\limits^1_0\frac{\text{dt}}{\text{t}^2}$
$=\Big[\log_{\text{e}}|\text{t}|\Big]^2_1$
$=\log_{\text{e}}2-\log_{\text{e}}1$
$=\log_{\text{e}}2-0$
$=\log_{\text{e}}2$

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Question 1582 Marks

Evaluate the definite integral in Exercise:
$\int_{2}^{3}\frac{1}{\text{x}}\text{dx}$

Answer

$\text{Let} \ \text{I}=\int\limits_{2}^{3}\frac{1}{\text{x}}\ \text{dx}$$\int\frac{1}{\text{x}}\text{dx}=\text{log}|\text{x|}=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(3)-\text{F}(2)$
$=\text{log|3|}-\text{log|2|}=\text{log}\frac{3}{2}$

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Question 1592 Marks

Evaluate the definite integral in Exercise:$\int\limits_0^{\frac{\pi}{4}}\sin2\text{x}\ \text{dx}$

Answer

$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\text{x}}{4}}\sin2\text{x} \ \text{dx}$$\int\sin2\text{x}\ \text{dx}=\bigg(\frac{-\cos2\text{x}}{2}\bigg)=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}\bigg(\frac{\pi}{4}\bigg)-\text{F}(0)$
$=-\frac{1}{2}\pi\bigg[\cos2\bigg(\frac{\pi}{4}\bigg)-\cos0\bigg] $
$=-\frac{1}{2}\bigg[\cos\bigg(\frac{\pi}{2}\bigg)-\cos0\bigg]$
$=-\frac{1}{2}[0-1]$
$=\frac{1}{2}$

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Question 1602 Marks

Evaluate the definite integral in Exercise:
$\int^{\pi}_{0}(\sin^{2}\frac{\text{x}}{2}-\cos^{2}\frac{\text{x}}{2})\text{dx}$

Answer

$\text{Let}\text{I}=\int\limits_{0}^{\pi}\bigg(\sin^{2}\frac{\text{x}}{2}-\cos^{2}\frac{\text{x}}{2}\bigg)\text{dx}$$=-\int\limits_{0}^{\pi}\bigg(\cos^{2}\frac{\text{x}}{2}-\sin^{2}\frac{\text{x}}{2}\bigg)\text{dx}$
$=-\int\limits_{0}^{\pi}\cos\text{x}\ \text{dx}$
$\int\cos\text{x}\text{dx}=\sin\text{x}=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(\pi)-\text{F}(0)$
$=\sin\pi-\sin0$
$=0$

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Question 1612 Marks

Integrate the function in Exercise:
$\frac{\cos\text{x}}{\sqrt{4-\sin^{2}}\text{x}}$

Answer

$\frac{\cos\text{x}}{\sqrt{4-\sin^{2}\text{x}}}$$\text{Let}\ \sin\text{x}=\text{t}\Rightarrow\cos\text{x}\ \text{dx}=\text{dt}$
$\Rightarrow\int\frac{\cos\text{x}}{\sqrt{4-\sin^{2}\text{x}}}\text{dx}=\int\frac{\text{dt}}{\sqrt{(2)^{2}-\text{(t)}^{2}}}$
$=\sin^{-1}\bigg(\frac{\text{t}}{2}\bigg)+\text{C}$
$=\sin^{-1}\bigg(\frac{\sin\text{X}}{2}\bigg)+\text{C}$

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Question 1622 Marks

Integrate the function in Exercise:
$\frac{\text{e}^{5\log\text{x}}-\text{e}^{4\log\text{x}}}{\text{e}^{3\log\text{x}}-\text{e}^{2\log\text{x}}}$

Answer

$\frac{\text{e}^{5\log\text{x}}-\text{e}^{4\log\text{x}}}{\text{e}^{3\log\text{x}}-\text{e}^{2\log\text{x}}}=\frac{\text{e}^{4\log\text{x}}\big(\text{e}^{\log\text{x}}-1\big)}{\text{e}^{2\log\text{x}}\big(\text{e}^{\log\text{x}}-1\big)}$$=\text{e}^{2\log\text{x}}$
$=\text{e}^{\log\text{x}^{2}}$
$=\text{x}^{2}$
$\therefore\int\frac{\text{e}^{5\log\text{x}}-\text{e}^{4\log\text{x}}}{\text{e}^{3\log\text{x}}-\text{e}^{2\log\text{x}}}\text{dx}=\int\text{x}^{2}\text{dx}=\frac{\text{x}^{3}}{3}+\text{C}$

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Question 1632 Marks

Evaluate the following:
$\int\frac{(\text{x}^2+2)}{\text{x}+1}\text{dx}$

Answer

Let $\text{I}=\int\frac{\text{x}^2+2}{\text{x}+1}\text{dx}$$=\int\Big(\text{x}-1+\frac{3}{\text{x}+1}\Big)\text{dx}$
$=\int(\text{x}-1)\text{dx}+3\int\frac{1}{\text{x}+1}\text{dx}$
$=\frac{\text{x}^2}{2}-\text{x}+3\log\big|(\text{x}+1)\big|+\text{C}$

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MATHS 2 Marks Questions