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Question

Mathematics Question on integral

Evaluate the definite integral: ∫01(xex+sinπx4)dx∫_0^1(xe^x+sin\frac{πx}{4})dx

Answer

Let I=∫01(xex+sinπx4)dx∫_0^1(xe^x+sin\frac{πx}{4})dx

∫(xex+sinπx4)dx=x∫exdx−∫(ddx)∫exdxdx+−cosπx4π4∫(xe^x+sin\frac{πx}{4})dx=x∫e^xdx-∫{(\frac{d}{dx})∫e^xdx}dx+{-\frac{cos\frac{πx}{4}}{\frac{π}{4}}}

=xex−∫exdx−4ππcosx4=xe^x-∫e^xdx-\frac{4π}{π}cos\frac{x}{4}

=xex−ex−4ππcosx4=xe^x-e^x-\frac{4π}{π}cos\frac{x}{4}

=F(x)=F(x)

By second fundamental theorem of calculus,we obtain

I=F(1)−F(0)I=F(1)-F(0)

=(1.e1−e1−4πcosπ4)−(0.e°−e°−4πcos0)=(1.e^1-e^1-\frac{4}{π}cos\frac{π}{4})-(0.e°-e°-\frac{4}{π}cos0)

=e−e−4π(1√2)+1+4π=e-e-\frac{4}{π}(\frac{1}{√2})+1+\frac{4}{π}

=1+4π−2√2π=1+\frac{4}{π}-\frac{2√2}{π}