Definite Integration
Integral of Piecewise Exponential Function
nta_pyq_2023_apr
Grade 12
Question:
Let $f:[0,2]\to\mathbb{R}$ be defined as $f(x)=\begin{cases}e^{\min\{x^2,x-[x]\}}, & x\in[0,1)\\e^{[x-\log_e x]}, & x\in[1,2]\end{cases}$. Then $\displaystyle\int_0^2 x f(x)\,dx$ is
1+\dfrac{3e}{2}
(e-1)\!\left(e^2+\dfrac{1}{2}\right)
2e-1
2e-\dfrac{1}{2}
Step-by-Step Solution
Key Concept: On $[0,1)$: $x^2<\{x\}$ for $x\in(0,1)$, so $f(x)=e^{x^2}$. On $[1,2]$: $x-\ln x\in[1,2)$ so $[x-\ln x]=1$ and $f(x)=e$.
$\int_0^1 xe^{x^2}dx=\frac{e-1}{2}$; $\int_1^2 ex\,dx=\frac{3e}{2}$. Total $=2e-\frac{1}{2}$.
Correct Answer: 4