## Integration Worksheet6 solution4

In this page integration worksheet6 solution4 we are going to see solution of some practice question from the worksheet of integration.

Question 9

Integrate the following with respect to x,   2 x e ^(3 x)

Solution:

Here we are going to use the method partial differentiation to integrate the given question.

∫ 2 x e ^(3 x) dx

u = x             dv = e ^(3 x)

du = dx           v = e^(3x)/3

= 2 {x [e^(3x)/3] - ∫[e^(3x)/3] dx}

= 2{(x/3) [e^(3x)] - (1/3)∫[e^(3x)] dx}

= (2x/3) [e^(3x)] - (2/3)[e^(3x)/3] + C

= (2x/3) [e^(3x)] - (2/9)[e^(3x)] + C

= (2/3) e^(3x)[x - (1/3)] + C

Question 10

Integrate the following with respect to x,  x² e ^(2 x)

Solution:

Here we are going to use the method partial differentiation to integrate the given question.

∫ x² e ^(2 x) dx

u = x²                 dv = e ^(2 x)

du = 2 x dx           v = e^(2x)/2

= {x²e^(2x)/2 - ∫ [e^(2x)/2] 2 x dx}

= {(x²/2)e^(2x) - ∫ [x e^(2x) dx}

u = x            dv = e^(2x)

du = dx          v = e^(2x)/2

= x [e^(2x)/2] - ∫ [e^(2x)/2] dx

= (x/2) [e^(2x)] - (1/2)∫ [e^(2x)] dx

= (x/2) [e^(2x)] - (1/2)[e^(2x)/2] + C

= (x/2) [e^(2x)] - (1/4)[e^(2x)] + C

= (1/2) (e^(2x))[x - (1/2)] + C

Question 11

Integrate the following with respect to x,  x² cos 3 x

Solution:

Here we are going to use the method partial differentiation to integrate the given question.

∫ x² cos 3 x dx

u = x²                 dv = cos 3 x

du = 2 x dx           v = sin 3 x/3

= {x² sin 3x/3) - ∫ [sin 3 x/3] 2 x dx}

= (x²/3)sin 3x - (2/3)∫ x [sin 3 x] dx

u = x       dv = sin 3x

du = dx      v = -cos 3x/3

= (x²/3)sin 3x - (2/3){[x (-cos 3 x/3)] - ∫ [-cos 3x/3] dx}

= (x²/3)sin 3x - (2/3)[x (-cos 3 x/3)] - (2/3)∫ [cos 3x/3] dx

= (x²/3)sin 3x + (2/9)[x cos 3 x] - (2/9)∫ [cos 3x] dx

= (x²/3)sin 3x + (2/9)[x cos 3 x] - (2/27)[sin 3x] + C

integration worksheet6 solution4 integration worksheet6 solution4