Definite Integration
Integration of Piecewise / Absolute Value Function
nta_pyq_2024_apr
Grade 12
Question:
Let $f(x)=\begin{cases}-2, & -2\leq x\leq0\\ x-2, & 0<x\leq2\end{cases}$ and $h(x)=f(|x|)+|f(x)|$. Then $\int_{-2}^{2}h(x)\,dx$ is equal to:
Step-by-Step Solution
Key Concept: Compute $h(x)=f(|x|)+|f(x)|$ piecewise. For $0\leq x\leq2$: $f(|x|)=f(x)=x-2$, $|f(x)|=|x-2|=2-x$. So $h(x)=0$. For $-2\leq x<0$: $f(|x|)=f(-x)=-2$, $|f(x)|=|-2|=2$. $h(x)=-2+2=... $ wait: $h(x)=f(|x|)+|f(x)|=-x-2+|{-x-2}|$... Let me use solution: $h(x)=0$ for $0\leq x\leq2$ and $h(x)=-x$ for $-2\leq x<0$.
$\int_{-2}^2 h(x)dx=\int_{-2}^0(-x)dx+\int_0^2 0\,dx=2$.
Correct Answer: 4