Mathematics

# Solve : $\displaystyle \int _0^3 x^3+2x+8 dx$

$\dfrac {213}4$

##### SOLUTION
consider, $I=\displaystyle \int_0^3 (x^3 + 2x + 3)dx$

$=\displaystyle \int_0^3 x^3dx + \int_0^3 2xdx + \int_0^38dx$

$=\left[\dfrac{x^4}{4}\right]^3_0 + [x^2]^3_0 + [8x]^3_0$

$=\dfrac{81}{4} + 9 + 24 = \dfrac{213}{4}$

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

#### Realted Questions

Q1 Single Correct Medium
Resolve into partial fraction $\displaystyle \frac{x^3-3x-2}{(x^2+x+1)(x+1)^2}$
• A. $\displaystyle \frac{3x-1}{x^2+x+1}+\frac{2}{(x+1)^2}+\frac{3}{(x+1)}$
• B. $\displaystyle \frac{3x}{x^2+x+1}+\frac{2}{(x+1)^2}-\frac{3}{(x+1)}$
• C. $\displaystyle \frac{x-1}{x^2+x+1}+\frac{2}{(x+1)^2}-\frac{3}{(x+1)}$
• D. $\displaystyle \frac{3x-1}{x^2+x+1}+\frac{2}{(x+1)^2}-\frac{3}{(x+1)}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate : $\displaystyle \int _{ -2 }^{ 2 }{ \frac { { x }^{ 2 } }{ 1+{ 5 }^{ x } } dx } .$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Solve:$\displaystyle \int\dfrac{1}{\sqrt{1+4x^2}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int _{ \pi /4 }^{ 3\pi /4 }{ \dfrac { dx }{ 1+\cos { x } } }$ is equal to:
• A. $2$
• B. $-2$
• C. $-\dfrac {1}{2}$
• D. $\dfrac {1}{2}$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$