Mathematics

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

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

Let $x+2=t^{2}$.

Differentiating with respect to $x$, we get

$dx=2t dt$

Also $x=0\Rightarrow t^{2}=2\Rightarrow t=\sqrt{2}$ and, $x=2\Rightarrow t^{2}=4\Rightarrow t=2$

So, lower limit is $\sqrt 2$ and upper limit is $2$

$\therefore I=\displaystyle\int_{\sqrt{2}}^{2}(t^{2}-2)\sqrt{t^{2}} \ \cdot 2t dt=2\displaystyle\int_{\sqrt{2}}^{2}(t^{4}-2t^{2})dt=2\left[\dfrac{t^{5}}{5}-\dfrac{2t^{3}}{3}\right]_{\sqrt{2}}^{2}$

$\Rightarrow I=2\left[\left(\dfrac{32}{5}-\dfrac{16}{3}\right)-\left(\dfrac{4\sqrt{2}}{5}-\dfrac{4\sqrt{2}}{3}\right)\right]=2\left(\dfrac{16}{15}+\dfrac{8\sqrt{2}}{15}\right)=\dfrac{32+16\sqrt{2}}{15}$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int \sqrt{\dfrac{a+x}{a-x}}dx$ is equal to-
• A. $a\cos^{-1} (x/a)-\sqrt{a^{2}-x^{2}}+c$
• B. $a\sin^{-1} (x/a)-\sqrt{a^{2}+x^{2}}+c$
• C. $a\cos^{-1} (x/a)-\sqrt{a^{2}+x^{2}}+c$
• D. $a\sin^{-1} (x/a)-\sqrt{a^{2}-x^{2}}+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evaluate: $\underset {n \rightarrow \infty} \lim \displaystyle \sum_{r=0}^{n-1}\frac{1}{n+r}$
• A. $2 \log 2$
• B. $\dfrac{1}{2} \log 2$
• C. $\dfrac{1}{4} \log 2$
• D. $\log 2$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
$\displaystyle\int_{0}^{\pi/2} \sin x \cos x\ dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the following integrals
$\int { \cfrac { 1 }{ \sqrt { 3 } \sin { x } +\cos { x } } } dx$

$\int {\dfrac{{1 - \frac{1}{{{x^2}}}}}{{{x^2} + \dfrac{1}{{{x^2}}} + 3}}} \,\,dx$