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# Integration Maths Help?

Not getting the correct answer with this question. Can anyone help? greatly appreciated

### 4 Answers

- husoskiLv 71 month agoFavorite Answer
You can get the correct answer from Wolfram Alpha, if that's all you needed. (Yes, it will do "area between" computation, but you may need to play with it a bit to find out how it expects you to ask the question.

To solve (a), I'd integrate the terms separately. The integral of -1 dx from x=2 to 4 is clearly -2. It's a rectangle.

For the 2/(x-1) term substitute u=x-1, du = dx. The integral becomes du/u integrated from u=1 to 3. That's ln(3) - ln(1) = ln(3)

Put both terms together to get ln(3) - 2 as the answer for part a.

For part (b) there's more work because f(x) = 2/(x-1) - 1 goes from positive to negative at x=3. You need to add the absolute values of the integrals from 2 to 3 and from 3 to 4, and add the absolute values. The procedure for each of those integrals is the same as above, with u=x-1 each time.

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- PhilipLv 61 month ago
(a) J = Int'l over(2,4) of [2/(x-1) -1]dx, x =/= 1. Put u=(x-1). Then dx=du &

(b) J = Int'l over(1,3) of [2/u -1]du = f(3)-f(1), where f(u) = 2lnu - u.

f(3) = 2ln3 -3, f(1) = 2ln1 -1 = -1 and f(3)-f(1) = 2ln3 -2 = 2(ln3 -1) =

o.1972245773.

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- 1 month ago
Remember, int((f(x) + g(x)) * dx) = int(f(x) * dx) + int(g(x) * dx)

int(2 * dx / (x - 1)) - int(1 * dx)

u = x - 1

du = dx

int(2 * du / u) - int(dx)

Integrate

2 * ln|u| - x + C =>

2 * ln|x - 1| - x + C

Now, before we go crazy and just plug in our domain endpoints, we need to determine that 2 / (x - 1) was both continuous and differentiable along the domain. There's a discontinuity when x = 1, and that's it, so we're good. Just plug in your domain endpoints now

2 * ln|4 - 1| - 4 - 2 * ln|2 - 1| + 2 =>

2 * (ln|3| - ln|1|) - 2 =>

2 * (ln(3) - 0) - 2 =>

2 * ln(3) - 2 =>

2 * (ln(3) - 1)

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- BryceLv 71 month ago
The function goes below the x-axis at x= 3, so you have to make two calculations: one on the interval 2≤ x ≤3 and another on 3≤ x ≤4. Take the negative of the second calculation because y is below the x-axis.

2ln|x - 1| - x= ln4 - 3 - (-2) - (ln9 - 4)

= ln(4/9) + 3≈ 2.189069784

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