Question 512 Marks
Evaluate the following integrals:
$\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
$\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
Answer
View full question & answer→We have,
$\text{I}=\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
We know that,
$\text{x}[\text{x}]=\begin{cases}\text{x}\times0,&0<\text{x}<1\\\text{x}\times1,&1<\text{x}<2\end{cases}$
i.e.,
$\text{x}[\text{x}]=\begin{cases}0,&0<\text{x}<1\\\text{x},&1<\text{x}<2\end{cases}$
$\therefore\ \text{I}=\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
$=\int\limits^{1}_0\text{x}[\text{x}]\text{dx}+\int\limits^{2}_1\text{x}[\text{x}]\text{dx}$
$=\int\limits^{1}_0{0}\text{ dx}+\int\limits^{2}_1(\text{x})\text{dx}$
$=0+\Big[\frac{\text{x}^2}{2}\Big]^2_1$
$=\frac{2^2}{2}-\frac{1^2}{2}$
$=\frac{3}{2}$
$\text{I}=\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
We know that,
$\text{x}[\text{x}]=\begin{cases}\text{x}\times0,&0<\text{x}<1\\\text{x}\times1,&1<\text{x}<2\end{cases}$
i.e.,
$\text{x}[\text{x}]=\begin{cases}0,&0<\text{x}<1\\\text{x},&1<\text{x}<2\end{cases}$
$\therefore\ \text{I}=\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
$=\int\limits^{1}_0\text{x}[\text{x}]\text{dx}+\int\limits^{2}_1\text{x}[\text{x}]\text{dx}$
$=\int\limits^{1}_0{0}\text{ dx}+\int\limits^{2}_1(\text{x})\text{dx}$
$=0+\Big[\frac{\text{x}^2}{2}\Big]^2_1$
$=\frac{2^2}{2}-\frac{1^2}{2}$
$=\frac{3}{2}$