a)
I=∫21(2x2+7x+4)2dx
Expand the parentheses to obtain:
I=∫214x4+28x3+65x2+56x+16dx=[45x5+7x4+653x3+28x2+16x]21=381.4666666667=381.47 to 2 decimal places
b)
I=∫1501x+1dx=[ln(x+1)]150=ln(16)−ln(1)=ln(161)=2.77 to 2 decimal places
c)
I=∫π0xsin(8x)dx
We use integration by parts.
Recall that:
∫udvdxdx=uv−∫dudxvdx
Here we set u=x and dvdx=sin(8x)
Hence v=−18cos(8x)
So I=[−x8cos(8x)]π0−∫π0−18cos(8x)dx=−π8−[−164sin(8x)]π0=−0.39 to 2 decimal places
d)
I=∫20x2e−2xdx
Use integration by parts twice with u=x2 and dvdx=e−2x⇒v=−12e−2x
I=[−x22e−2x]20+∫20xe−2xdx=−2e−4+1([−x2e−2x]20+12∫20e−2xdx)=−2e−4+1(−e−4+12[−12e−2x]20)=−3e−4−14(e−4−1)=0.1905 to 4 decimal places