Let f(x)=\ array lr -2, & -2 ≤ x ≤ 0 x-2, & 0 < x ≤ 2 array . and...

Let

f(x)=\left\{

\begin{array}{lr}-2, & -2 \leq x \leq 0 \\ x-2, & 0 < x \leq 2\end{array}

\right.

and

h(x)=f(|x|)+|f(x)|

. Then

\int_{-2}^2 h(x) \mathrm{d} x

is equal to : $Let f(x)=\ array lr -2, & -2 ≤ x ≤ 0 x-2, & 0 < x ≤ 2 array . and h(x)=f(|x|)+|f(x)| . Then ∫_ -2 ^2 h(x) d x is equal$ $Let f(x)=\ array lr -2, & -2 ≤ x ≤ 0 x-2, & 0 < x ≤ 2 array . and h(x)=f(|x|)+|f(x)| . Then ∫_ -2 ^2 h(x) d x is equal$ $Let f(x)=\ array lr -2, & -2 ≤ x ≤ 0 x-2, & 0 < x ≤ 2 array . and h(x)=f(|x|)+|f(x)| . Then ∫_ -2 ^2 h(x) d x is equal$

Options

Correct answer

D. 2

Step-by-step solution

$Let f(x)=\ array lr -2, & -2 ≤ x ≤ 0 x-2, & 0 < x ≤ 2 array . and h(x)=f(|x|)+|f(x)| . Then ∫_ -2 ^2 h(x) d x is equal$
$Let f(x)=\ array lr -2, & -2 ≤ x ≤ 0 x-2, & 0 < x ≤ 2 array . and h(x)=f(|x|)+|f(x)| . Then ∫_ -2 ^2 h(x) d x is equal$

h(x)=\left\{

\begin{array}{cc}x-2+2-x=0, & 0 \leq x \leq 2 \\ -x-2+2=-x & -2 \leq x < 0\end{array}

\right.

$Let f(x)=\ array lr -2, & -2 ≤ x ≤ 0 x-2, & 0 < x ≤ 2 array . and h(x)=f(|x|)+|f(x)| . Then ∫_ -2 ^2 h(x) d x is equal$

\Rightarrow \int_0^2 \mathrm{~h}(\mathrm{x}) \mathrm{dx}=0

and

\int_{-2}^0 \mathrm{~h}(\mathrm{x}) \mathrm{dx}=2

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