Integration question?

Question

part 1
By using the substitution u= sqroot(x+3), find the integration of (e^sqroot(x+3))/sqroot(x+3) wrt x.

part 2
Hence evaluate the integration of e^sqroot(x+3) wrt x, with integrals 0 to 1, giving ur answer in exact form.

Part 2 is the only part i need help in. so any help, even just by pointing me in the right direction would be appreciated. thanks

Answer 1

You must remember to back substitute the value of x. As the person above me answered, the answer to 1 is 2*e^(u)=2*e^(sqrt(x+3)).
No plug in in the limits to get
2*e^(sqrt(1+3))-2*e^(sqrt(0+3))
=2*e^(2)-2*e^(sqrt(3))
=2*[e^(2)-e^(sqrt(3))]

Answer 2

Part 1 has an expression with 2 terms and part 2 has an expression with one term. What does this make you think of?..... Yes, you are right. The product rule of differentiation.

d(uv) = udv + vdu

You may carry on from here if you like. I have worked it out further below but I tend to make little errors in the details although I do try to be thorough in checking.

Make the integral from above into the udv term so:
u = e^(SQRT(x + 3))
dv = dx/(SQRT(x + 3))

And udv = e^(SQRT(x + 3)) dx / (SQRT(x + 3)) ... the equation from part 1 and whose integral is 2 e^(SQRT(x + 3))

du = e^(SQRT(x + 3)) /(2 SQRT(x + 3))
v = 2(SQRT(x + 3))

And vdu = e^(SQRT(x + 3)) ... the expression in part 2

Reorder so the left hand side is the expression you want to integrate in part 2.

vdu = d(uv) - udv

Integrate the right hand side. The first term is just uv and the second is the integral from part 1. I'm going to call the integral of udv F so:

F = 2(SQRT(x + 3))e^(SQRT(x + 3)) - 2e^(SQRT(x + 3))
F = 2e^(SQRT(x + 3)) (SQRT(x + 3) - 1) evaluated from 0 to 1
F = 2e^2 - 2(SQRT(3) - 1)e^(SQRT(3))

Answer 3

Since u=sqrt(x+3), du=(1/2)*(x+3)^(-1/2) dx
du=(1/2)*dx/u
Then dx=2*u*du
Then the integral becomes
int(2*exp(u)*du)
=2*exp(u)

for the evaluation in part 2,
2*(exp(1)-exp(0))=2*(e-1)