1
Question
Evaluate ∫2ππsinxdx
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Solution
Apply,∫baf(x)dx=f(b)−f(a)=limx−b−(f(x))−limx−a+(f(x))
Hence,∫sinxdx=−cosx+c
limx→π+(−cos(x))=1
limx→2π−(−cos(x))=−1
=−1−1
=−2
∫baf(x)dx=f(b)−f(a)=limx−b−(f(x))−limx−a+(f(x))
Hence,
∫sinxdx=−cosx+c
limx→π+(−cos(x))=1
limx→2π−(−cos(x))=−1
=−1−1
=−2
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