Mathematics

# Evaluate the integral$\displaystyle \int_{0}^{a}f(x)dx+\int_{0}^{a}f(2a-x)dx$

$\displaystyle \int_{0}^{2a}f(x)dx$

##### SOLUTION
$I=\displaystyle\int_{0}^{a}f(x)dx+\int_{0}^{a}f(2a-x)dx$

Using the property $\displaystyle\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ in the second part:-

$\implies I=\displaystyle\int_{0}^{a}f(x)dx+\int_{0}^{a}f\left(2a-(a-x)\right)dx$

$\implies I=\displaystyle\int_{0}^{a}f(x)dx+\int_{0}^{a}f(a+x)dx$

$\implies I=\displaystyle\int_{0}^{a}f(x)dx+\int_{a}^{2a}f(x)dx$

$=\displaystyle\int_{0}^{2a}f(x)dx$

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

#### Realted Questions

Q1 One Word Medium
Evaluate $\displaystyle \int \frac{dx}{\left ( a^{2}+x^{2} \right )^{3/2}}$
$\displaystyle I=\frac{x}{a^{2}\left ( x^{2}+a^{2} \right )^{K}}+c$
What is K?

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate : $\displaystyle \int { { e }^{ x }\left[ \cfrac { 1 }{ 1-x } +\cfrac { 1 }{ { \left( 1-x \right) }^{ 2 } } \right] } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle \int\frac{e^{4x}-1}{e^{2x}}\log(\frac{e^{2x}+1}{e^{2x}-1})ck=\frac{t^{2}}{2}\log t-\frac{t^{2}}{4}-\frac{u^{2}}{2} \log u + \frac{u^{2}}{4}+c$ then
• A. $\displaystyle t=e^{-x}-e^{x},u=e^{x}+e^{-x}$
• B. $\displaystyle t=e^{x}-e^{-x},u=e^{x}+e^{-x}$
• C. $\displaystyle u=e^{-x}-e^{x},t=e^{x}+e^{-x}$
• D. $\displaystyle t=e^{x}+e^{-x},u=e^{x}-e^{-x}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the following definite integral:
$\displaystyle \int_{-1}^{1}\dfrac {1}{x^2 +2x+5}dx$

$2x^2e^{x^2}$