help with integral question pls!! (1 Viewer)

[miracu]

Can someone show their long working out for this?



I keep getting 34b^3/3 as an answer. I know there's a quicker method to answering this but I'm only in 2 unit and we haven't learnt it.

 

[InteGrand]

Can someone show their long working out for this?

View attachment 33943

I keep getting 34b^3/3 as an answer. I know there's a quicker method to answering this but I'm only in 2 unit and we haven't learnt it.

Expand the numerator using the formula for (A + B)^2, and then simplify by splitting the fraction. This will leave us with terms that can be integrated using the power rule.

 

[miracu]

Expand the numerator using the formula for (A + B)^2, and then simplify by splitting the fraction. This will leave us with terms that can be integrated using the power rule.

I did that, it's the part where I substitute 4b^2 and b^2 which seems to be the problem- I just want to see the full working out to see where I went wrong

 

[InteGrand]

I did that, it's the part where I substitute 4b^2 and b^2 which seems to be the problem- I just want to see the full working out to see where I went wrong

Why not post your working, and maybe someone can spot your mistake. In writing out your working here, it's possible that you may spot a mistake too.

 

[pikachu975]

Can someone show their long working out for this?

View attachment 33943

I keep getting 34b^3/3 as an answer. I know there's a quicker method to answering this but I'm only in 2 unit and we haven't learnt it.

Integrand becomes (x+2bsqrt(x) + b^2)/sqrt(x)
= sqrt(x) + 2b + b^2 * x^(-1/2)

Integrate
= 2/3 * x^3/2 + 2bx + 2b^2 * sqrt(x)

Sub in boundaries
= 2/3 * (4b^2)^3/2 + 8b^3 + 4b^3 - 2/3 * b^3 - 2b^3 - 2b^3
= 16/3 * b^3 + 12b^3 - 2/3 b^3 - 4b^3
= 38b^3 / 3