# Thread: Simple Integral with U substitution

1. ## Simple Integral with U substitution

Hi,

I'm trying to solve this integral

$\displaystyle \int_{0}^{0.4}12.5x(1-x)^4 dx$

I'm letting u = 1-x, so that I have

$\displaystyle \int_{0.6}^{1}12.5(1-u)u^4 du$

Then

$\displaystyle 2.5u^5 - \frac{25}{12}u^6 |^{1}_{0.6} =0.3195$

But if I replace u with 1-x then

$\displaystyle 2.5(1-x)^5 - \frac{25}{12}(1-x)^6 |^{0.4}_{0}$

I don't get the correct answer.

Why is this? Can I not substitute (1-x) for u?

Thanks.

2. ## Re: Simple Integral with U substitution

Originally Posted by downthesun01
Hi,

I'm trying to solve this integral

$\displaystyle \int_{0}^{0.4}12.5x(1-x)^4 dx$

I'm letting u = 1-x, so that I have

$\displaystyle \int_{0.6}^{1}12.5(1-u)u^4 du$

Then

$\displaystyle 2.5u^5 - \frac{25}{12}u^6 |^{1}_{0.6} =0.3195$

But if I replace u with 1-x then

$\displaystyle 2.5(1-x)^5 - \frac{25}{12}(1-x)^6 |^{0.4}_{0}$

I don't get the correct answer.

Why is this? Can I not substitute (1-x) for u?

Thanks.

Let's go through the substitution a little more carefully:

$$\int_0^{0.4} 12.5x(1-x)^4dx$$

$$u = 1-x, du = -dx$$

So:

$$\int_0^{0.4} 12.5x(1-x)^4dx = -\int_{1}^{0.6} 12.5(1-u)u^4du$$

You used the fact that:

$$-\int_1^{0.6}f(u)du = \int_{0.6}^1 f(u)du$$

But, when you switch back, you have to remember you reversed the bounds of integration. So, you need:

$$\left. 2.5(1-x)^5 - \dfrac{25}{12}(1-x)^6 \right|_{0.4}^0$$

This will give you the same answer.

3. ## Re: Simple Integral with U substitution

I knew it was something obvious that I just wasn't seeing. Thank you.