Mathematics

# If $f(x)=\dfrac{4}{\pi}\sin\left(\dfrac{\pi}{2}x\right)+B$ and $\int^0_1f(x)dx=\cfrac{4}{\pi}\int\sin(\cfrac{\pi}{2}x)+B dx$, Find $B$

##### SOLUTION
$f(x)=\cfrac{4}{\pi}\sin(\cfrac{\pi}{2}x)+B$
$\int^0_1f(x)dx=\cfrac{4}{\pi}\int\sin(\cfrac{\pi}{2}x)+B dx$
$\implies \cfrac{\pi}{4}\int^1_0f(x)dx=\cfrac{-2}{\pi}[\cos \cfrac{\pi}{2}x+Bx]^0_1$
$\implies \cfrac{2}{\pi}=\cfrac{-2}{\pi}[0-1] +B$
$\implies \cfrac{2}{\pi}=\cfrac{2}{\pi}+B\implies B=0$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
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