Mathematics

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

SOLUTION

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

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

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

$\Rightarrow t=\sqrt{2}$

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

$\Rightarrow t=2$

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

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

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

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

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

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

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

Realted Questions

Q1 Single Correct Medium
Evaluate: $\displaystyle \int\cfrac{11^{\tfrac{x}{2}}}{\sqrt{11^{-x}+11^{x}}}dx$
• A. $\displaystyle \frac{1}{\log_{11}e} \sin^{-1} 11^{x}+c$
• B. $\displaystyle \frac{\cosh^{-1}11^{x}}{\log 11}+c$
• C. $\log_{11}e\log|11^{x}+\sqrt{11^{2x}-1}|+c$
• D. $\log_{11}e.\log[11^{x}+\sqrt{1+11^{2x}}]+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The value of $\displaystyle \int_{0}^{\pi /2}\displaystyle \frac{\sin 8x\log \left ( \cot x \right )}{\cos 2x}dx$ is
• A. $\pi$
• B. $\dfrac {5\pi}{2}$
• C. $\dfrac {3\pi}{2}$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
The value of $\displaystyle\int 5^{5^{5^x}}\cdot 5^{5^x}\cdot 5^x dx$ is equal to?
• A. $\dfrac{5^{5^x}}{(ln 5)^3}+C$
• B. $5^{5^{5^x}}(ln 5)^3+C$
• C. $\dfrac{5^{5^{5^x}}}{(ln 5)^2}+C$
• D. $\dfrac{5^{5^{5^x}}}{(ln 5)^3}+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the given integral.
$\int { { x }^{ 3 }\log { x } } dx$

Evaluate $\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}}dx$