# Find a and b

1. Jun 25, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
If $\displaystyle \int^b_a \dfrac{x^n dx}{x^n + (16-x)^n} = 6$ and a+b=16, then find a and b.

3. The attempt at a solution

$\displaystyle \int^b_a \dfrac{dx}{1 + (16/x - 1)^n} = 6$

I tried substitution but it did not work.

2. Jun 25, 2013

### Curious3141

What sub did you try? Hint: try $x = 16-y$ on the original integral.

This is a simple algebra problem, more or less. Very little actual integration need be done.

3. Jun 25, 2013

### pasmith

Hint:
$$\int_a^b\frac{x^n}{x^n + (16-x)^n}\,\mathrm{d}x = \int_a^b 1\,\mathrm{d}x - \int_a^b\frac{(16-x)^n}{x^n + (16-x)^n}\,\mathrm{d}x$$
Can you find a substitution which turns the integrand of the second integral on the right into the integrand of the integral on the left?

4. Jun 26, 2013

### sankalpmittal

Why do you need substitution right now ?

Apply the property:

I= $\displaystyle \int^b_a \dfrac{x^n dx}{x^n + (16-x)^n}$

And I= $\displaystyle \int^b_a \dfrac{(16-x)^n dx}{(16-x)^n + (16-(16-x))^n}$

Now add the two to get,

2I= ab∫dx

Can you proceed? You already know that I=6...

5. Jun 26, 2013

### utkarshakash

Oh that was so easy. I first thought of applying property but somehow couldn't notice that a+b=16 was already given.