Mathematics

# Solve:$\displaystyle\int_{0}^{2}x\sqrt{x+2}dx$

##### SOLUTION

$I=\displaystyle\int_{0}^{2}x\sqrt{x+2}dx$

Let $x+2=t^{2}$. then , $dx=2t dt$

Also $x=0\Rightarrow t^{2}=2$

$\Rightarrow t=\sqrt{2}$

$x=2\Rightarrow t^{2}=4$

$\Rightarrow t=2$

$\therefore I=\displaystyle\int_{\sqrt{2}}^{2}(t^{2}-2)\sqrt{t^{2}}2t dt$

$=2\displaystyle\int_{\sqrt{2}}^{2}(t^{4}-2t^{2})dt$

$=2\left[\dfrac{t^{5}}{5}-\dfrac{2t^{3}}{3}\right]_{\sqrt{2}}^{2}$

$\Rightarrow I=2\left[\left(\dfrac{32}{5}-\dfrac{16}{3}\right)-\left(\dfrac{4\sqrt{2}}{5}-\dfrac{4\sqrt{2}}{3}\right)\right]$

$=2\left(\dfrac{16}{15}+\dfrac{8\sqrt{2}}{15}\right)$

$=\dfrac{32+16\sqrt{2}}{15}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

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• A. $1$
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• C. None of these
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q2 One Word Medium
Solve $\displaystyle\int \dfrac{ { e }^{ 2x }-{ e }^{ -2x } }{ { e }^{ 2x }+{ e }^{ -2x } } dx$

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Q3 Single Correct Hard
The value of $\displaystyle \int_{0}^{\pi /2} \dfrac {\sin 2t}{\sin^{4}t + \cos^{4}t} dt$
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• B. $\dfrac {\pi}{3}$
• C. $\dfrac {\pi}{4}$
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Q4 Single Correct Hard
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