Mathematics

# Solve the equation:-$\int_{0}^{2} [x^{2}-x+1]dx$

##### SOLUTION
$\displaystyle\int_{0}^{2}{\left({x}^{2}-x+1\right)dx}$

$=\left[\dfrac{{x}^{3}}{3}-\dfrac{{x}^{2}}{2}+x\right]_{0}^{2}$

$=\dfrac{1}{3}\left(8-0\right)-\dfrac{1}{2}\left(4-0\right)+\left(2-0\right)$

$=\dfrac{8}{3}-\dfrac{4}{2}+2$

$=\dfrac{8}{3}-2+2$

$=\dfrac{8}{3}$

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \mathrm{I}\mathrm{f}\frac{\mathrm{x}^{2}}{(\mathrm{x}^{2}+1)(\mathrm{x}^{2}+2)}=\frac{\mathrm{A}\mathrm{x}+\mathrm{B}}{\mathrm{x}^{2}+1}+\frac{\mathrm{C}\mathrm{x}+\mathrm{D}}{\mathrm{x}^{2}+2}$ then $(A,C)=$
• A. $(1,-1)$
• B. $(1,1)$
• C. $(1,2)$
• D. $(0,0)$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int\frac{(1+\log x)^{n}}{x}dx=$
• A. $\displaystyle (1+\log x)^{n}+c$
• B. $\displaystyle -\frac{(1+\log x)^{n+1}}{n+1}+c$
• C. $\displaystyle \frac{(1+\log x)^{n}}{n+1}+c$
• D. $\displaystyle \frac{(1+\log x)^{n+1}}{n+1}+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 One Word Hard
If ${I}_{1}=\displaystyle\int_{0}^{1}{\dfrac{{\tan}^{-1}{x}}{x}dx}$ and ${I}_{2}=\displaystyle\int_{0}^{1}{\dfrac{x}{\sin{x}}dx}$ then $\dfrac{{I}_{1}}{{I}_{2}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $I(m,n)=\int _{ 0 }^{ 1 }{ { t }^{ m }{ \left( 1-t \right) }^{ n }dt }$, then the expression for $I(m,n)$ in terms of $I(m+1,n-1)$ is
• A. $\cfrac{n}{m+1}$ $I(m+1,n-1)$
• B. $\cfrac{{2}^{n}}{m+1}-\cfrac{n}{m+1}$ $I(m+1,n-1)$
• C. $\cfrac{m}{m+1}$ $I(m+1,n-1)$
• D. $\cfrac{{2}^{n}}{m+1}+\cfrac{n}{m+1}$ $I(m+1,n-1)$

Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$