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Question Number 92772 by Power last updated on 09/May/20
Commented by mathmax by abdo last updated on 09/May/20
A=∫06[x]sin(πx6)dx⇒A=∑k=05∫kk+1ksin(πx6)dx=∑k=05k∫kk+1sin(πx6)dx=∑k=05k[−6πcos(πx6)]kk+1=−6π∑k=05k{cos(π(k+1)6)−cos(kπ6)}=−6π(cos(π3)−cos(π6)+cos(π2)−cos(π3)+cos(2π3)−cos(π2)+cos(5π6)−cos(2π3)+cos(π)−cos(5π6))=−6π{−32−12+12−1}=6π(1+32)=62π(1+3)
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