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
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