Question

5 mark for review
f(x)=\begin{cases}-1&\text{for }0leq x < 3\\1&\text{for }3 < x < 5\\3&\text{for }5leq xleq7end{cases}
let (f) be the piece - wise function given above. the value of (int_{0}^{7}f(x)dx) is
(a) 5
(b) 11
(c) 35
(d) nonexistent

Explanation:

Step1: Split the integral based on intervals

By the property of definite - integrals for piece - wise functions $\int_{0}^{7}f(x)dx=\int_{0}^{3}f(x)dx+\int_{3}^{5}f(x)dx+\int_{5}^{7}f(x)dx$.

Step2: Calculate $\int_{0}^{3}f(x)dx$

Since $f(x)= - 1$ for $0\leq x\lt3$, then $\int_{0}^{3}f(x)dx=\int_{0}^{3}-1dx=-x\big|_{0}^{3}=-(3 - 0)=-3$.

Step3: Calculate $\int_{3}^{5}f(x)dx$

Since $f(x)=1$ for $3\lt x\lt5$, then $\int_{3}^{5}f(x)dx=\int_{3}^{5}1dx=x\big|_{3}^{5}=5 - 3 = 2$.

Step4: Calculate $\int_{5}^{7}f(x)dx$

Since $f(x)=3$ for $5\leq x\leq7$, then $\int_{5}^{7}f(x)dx=\int_{5}^{7}3dx=3x\big|_{5}^{7}=3(7 - 5)=6$.

Step5: Sum up the results

$\int_{0}^{7}f(x)dx=-3 + 2+6=5$.

Answer:

A. 5