Solving an Integral Equation Involving x and 16

[utkarshakash]

Homework Statement

If \displaystyle \int^b_a \dfrac{x^n dx}{x^n + (16-x)^n} = 6 and a+b=16, then find a and b.

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.

 

[Curious3141]

Homework Statement

If \displaystyle \int^b_a \dfrac{x^n dx}{x^n + (16-x)^n} = 6 and a+b=16, then find a and b.

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.

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.

 

[pasmith]

Hint:
<br /> \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<br />
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?

 

[sankalpmittal]

Homework Statement

If \displaystyle \int^b_a \dfrac{x^n dx}{x^n + (16-x)^n} = 6 and a+b=16, then find a and b.

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.

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= _a^b∫dx

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

 

[utkarshakash]

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= _a^b∫dx

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

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