Mathematics

# Integrate $\displaystyle \int_{1}^{4}(x^2-x)dx$

##### SOLUTION
$\displaystyle \int_{1}^{4}(x^2-x)dx$

Using $\displaystyle\int{{x}^{n}dx}=\dfrac{{x}^{n+1}}{n+1}+C$, we get

$=\left[\dfrac{x^3}3-\dfrac {x^2}2\right]_1^4$

$=\dfrac{64}3-8-\dfrac 13+\dfrac 12$

$=\dfrac{27}2$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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