Question #91393
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
1
Expert's answer
2019-07-10T09:54:58-0400

For 0<x120<x\leq\frac{1}{2}


0xf(s)ds=0x1ds=s0x=x,\int_0^xf(s)ds=\int_0^x1ds=s\mid_0^x=x,

for 12<x1\frac{1}{2}<x\leq1


0xf(s)ds=0121ds+12x(1)ds=s012s12x=\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=

=12x+12=1x.=\frac{1}{2}-x+\frac{1}{2}=1-x.

Answer: b) 0xf(s)ds={x  0<x1/21x  1/2<x1\int_0^xf(s)ds=\begin{cases} x \text{, } \;0<x\leq1/2 \\ 1-x \text{, } \;1/2<x\leq1 \end{cases}


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