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
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