First slide

Evaluation of definite integrals

Question

If $f (x) = x | x |$ , then for any real number $a$ and $b$
with $a < b,$ the value of $\int_{a}^{b} f (x) d x$ equals

Moderate

A

$\frac{1}{3} (| b |^{3} - | a |^{3})$
B

$\frac{1}{3} |b^{3} - a^{3}|$
C

$\frac{1}{3} (a^{3} + b^{3})$
D

$\frac{1}{3} (a^{3} - b^{3})$

Solution

$\int_{a}^{b} f (x) d x = \int_{a}^{b} x | x | d x$
$= \{\begin{array}{cc} \int_{a}^{b} (- x^{2}) d x & b < 0 \\ \int_{a}^{0} - x^{2} d x + \int_{0}^{b} x^{2} d x, & a < 0 < b \\ \int_{a}^{b} x^{2} d x, & 0 < a \end{array}$
$= \{\begin{cases} - \frac{1}{3} (b^{3} - a^{3}), b < 0 \\ \frac{1}{3} (b^{3} + a^{3}), a < 0 < b \\ \frac{1}{3} (b^{3} - a^{3}), 0 < a \end{cases}$
$= \frac{1}{3} (| b |^{3} - | a |^{3})$

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