Mathematics

# Evaluate: $\displaystyle \int _0^1 (8x^2+16) dx$

##### SOLUTION
$I=\displaystyle \int _0^1 (8x^2+16)dx$

$I=8\displaystyle \int (x^2+2)dx$

$I=8 \left.\dfrac{x^3}3+16x\right|_0^1$

$I=\dfrac 83+16=\dfrac {56}3$

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

#### Realted Questions

Q1 Single Correct Medium
$\int_0^\infty {{x^n}{e^{ - x}}dx}$ (n is +ve integer) is equal to
• A. $(n - 1)!$
• B. $(n - 2)!$
• C. $(n + 1)!$
• D. $n!$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate:
$\displaystyle \int _{ 1 }^{ e }{ \left( \frac{1}{\sqrt{x \ln x}} + \sqrt{\frac{\ln x}{x}} \right) dx }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Calculate:
$\int (x^{3/2} - x^{1/2} + 5x) dx$.
• A. $\dfrac{2x^{5/2}}{3}-\dfrac{2x^{3/2}}{5}+\dfrac{5}{2}x^2+c$
• B. $\dfrac{2x^{5/2}}{5}-\dfrac{x^{3/2}}{4}+\dfrac{5}{2}x^2+c$
• C. $\dfrac{x^{5/2}}{5}-\dfrac{2x^{3/2}}{3}+\dfrac{5}{2}x^2+c$
• D. $\dfrac{2x^{5/2}}{5}-\dfrac{2x^{3/2}}{3}+\dfrac{5}{2}x^2+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
The value of $\displaystyle \lim_{n \rightarrow \infty} \Sigma_1^n \sin\left(\dfrac{\pi}{4} +\dfrac{\pi i}{2n} \right) \dfrac{\pi}{2n}=?$
• A. $\displaystyle \int_{\tfrac{\pi}{2}}^{\tfrac{\pi}{4}} \sin x dx$
• B. $\displaystyle \int_{\tfrac{\pi}{2}}^{\tfrac{3\pi}{4}} \sin x dx$
• C. $\displaystyle \int_\pi^{3\pi} \sin x dx$
• D. $\displaystyle \int_{\tfrac{\pi}{4}}^{\tfrac{3\pi}{4}} \sin x dx$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$