a)
I=∫61(6x2−9x+1)2dx
Expand the parentheses to obtain:
I=∫6136x4−108x3+93x2−18x+1dx=[365x5−27x4+31x3−9x2+x]61=27370=27370.00 to 2 decimal places
b)
I=∫1701x+6dx=[ln(x+6)]170=ln(23)−ln(6)=ln(236)=1.34 to 2 decimal places
c)
I=∫π0xsin(5x)dx
We use integration by parts.
Recall that:
∫udvdxdx=uv−∫dudxvdx
Here we set u=x and dvdx=sin(5x)
Hence v=−15cos(5x)
So I=[−x5cos(5x)]π0−∫π0−15cos(5x)dx=π5−[−125sin(5x)]π0=0.63 to 2 decimal places
d)
I=∫10x2e−3xdx
Use integration by parts twice with u=x2 and dvdx=e−3x⇒v=−13e−3x
I=[−x23e−3x]10+23∫10xe−3xdx=−e−3+23([−x3e−3x]10+13∫10e−3xdx)=−e−3+23(−e−3+13[−13e−3x]10)=−13e−3−29e−3−227(e−3−1)=0.0427 to 4 decimal places