How do you solve $\int_{0}^{2} x^{4} + 5 x d x$ $int_0^2 x^4+5x dx$ ?

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How do you solve $\int_{0}^{2} x^{4} + 5 x d x$ $int_0^2 x^4+5x dx$ ?

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Answer 1 · 2016-05-04T17:30:55

$\frac{82}{5}$ $82/5$

Explanation:

First, take the antiderivative:
${[\frac{1}{5} x^{5} + \frac{5}{2} x^{2}]}_{0}^{5}$ $[1/5x^5 + 5/2x^2]_0^2$

Then, evaluate:
$(\frac{1}{5} {(2)}^{5} + \frac{5}{2} {(2)}^{2}) - (0)$ $(1/5(2)^5 + 5/2(2)^2) - (0)$
$= \frac{32}{5} + 10$ $=32/5+10$
$= \frac{82}{5}$ $=82/5$

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How do you solve $\int_{0}^{2} x^{4} + 5 x d x$ $int_0^2 x^4+5x dx$ ?

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