Solved: "Evaluate the integral, if it exists. ∫03|x2−4|dx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

William Boggs

Answered question

2021-12-12

Evaluate the integral, if it exists.

\int_{0}^{3} | x^{2} - 4 | d x

Answer & Explanation

zesponderyd

Beginner2021-12-13Added 41 answers

Step 1
Given: An integral $\int_{0}^{3} | x^{2} - 4 | d x$
$x^{2} - 4 > 0$ , for $x \geq 2$
$x^{2} - 4 < 0$ for x<2
Thus, $\int_{0}^{3} | x^{2} - 4 | d x = \int_{0}^{2} - (x^{2} - 4) d x + \int_{2}^{3} (x^{2} - 4) d x$
Step 2
Now integrate the above integral with respect to x
$\int_{0}^{3} | x^{2} - 4 | d x = \int_{0}^{2} - (x^{2} - 4) d x + \int_{2}^{3} (x^{2} - 4) d x$
$= {[- \frac{x^{3}}{3} + 4 x]}_{0}^{2} + {[\frac{x^{3}}{3} - 4 x]}_{2}^{3}$
$= [- \frac{2^{3}}{3} + 4 (2) - (- \frac{0^{3}}{3} + 4 (0))] + [\frac{3^{3}}{3} - 4 (3) - (\frac{2^{3}}{3} - 4 (2))]$
$= [- \frac{8}{3} + 8] + [\frac{27}{3} - 12 - \frac{8}{3} + 8]$
$= [\frac{- 8 + 27 - 8}{3} + 4]$
$= \frac{11}{3} + 4$
$= \frac{11 + 12}{3}$
$= \frac{23}{3}$
Thus, the value of the definite integral $\int_{0}^{3} | x^{2} - 4 | d x$ is $\frac{23}{3}$

Lindsey Gamble

Beginner2021-12-14Added 38 answers

Given:

\int | x^{2} - 4 | d x

= \frac{x^{2} - 4}{| x^{2} - 4 |} \int (x^{2} - 4) d x

\int (x^{2} - 4) d x

Lets

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