How do you evaluate: $\int_{- 2}^{3} x^{2} - 3 d x$ $int_{-2}^3 x^2 - 3 dx$ ?

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How do you evaluate: $\int_{- 2}^{3} x^{2} - 3 d x$ $int_{-2}^3 x^2 - 3 dx$ ?

1 Answer

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Answer 1 · 2017-02-01T15:34:09

$- \frac{10}{3}$ $-10/3$

Explanation:

STEP 1: Take the antiderivative of the function
= ${[\frac{1}{3} x^{3} - 3 x]}_{- 2}^{3}$ $[1/3x^3 - 3x]_-2^3$

STEP 2: Evaluate between the founds
= $[(\frac{1}{3}) (3^{3}) - 3 (3)] - [(\frac{1}{3}) {(- 2)}^{3} - 3 (- 2)]$ $[(1/3) (3^3) - 3(3)] - [(1/3) (-2)^3 - 3(-2)]$

STEP 3: Compute your final answer
= $[9 - 9] - [- \frac{8}{3} + 6]$ $[9 - 9] - [-8/3 + 6]$
= $\frac{8}{3} - \frac{18}{3}$ $8/3 - 18/3$
= $- \frac{10}{3}$ $-10/3$

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How do you evaluate: $\int_{- 2}^{3} x^{2} - 3 d x$ $int_{-2}^3 x^2 - 3 dx$ ?

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Explanation:

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