Answer to Question #91393 in Calculus for Sajid

Question

Q. choose the correct option.
Q. If f(x)= 1, 0<x≤1/2. -1, ½<x≤1 then
a. ∫_0^x▒f(s)ds=x, 0<x≤1/2. -x, ½<x≤1
b. ∫_0^x▒f(s)ds=x, 0<x≤1/2. 1-x, ½<x≤1
c. ∫_0^x▒f(s)ds=1/2, 0<x≤1/2. 1, ½<x≤1
d. ∫_0^x▒f(s)ds=1-x, 0<x≤1/2. x, ½<x≤1

Answer 1

For $0<x\leq\frac{1}{2}$

$$\int_0^xf(s)ds=\int_0^x1ds=s\mid_0^x=x,$$

for $\frac{1}{2}<x\leq1$

$$\int_0^xf(s)ds=\int_0^{\frac{1}{2}}1ds+\int_{\frac{1}{2}}^x(-1)ds=s\mid_0^{\frac{1}{2}}-s\mid_{\frac{1}{2}}^x=$$

$$=\frac{1}{2}-x+\frac{1}{2}=1-x.$$

Answer: b) $\int_0^xf(s)ds=\begin{cases} x \text{, } \;0<x\leq1/2 \\ 1-x \text{, } \;1/2<x\leq1 \end{cases}$