# Solving an equation after integrating

1. Jun 30, 2011

### sara_87

1. The problem statement, all variables and given/known data

i have an equation that i want to solve for b:
$\int _0^1 (a+b x)^2 x^c dx=2$

2. Relevant equations
\int _0^1 (a+b x)^2 x^c dx=2
given: c>0

3. The attempt at a solution

To evaluate the integral on the left hand side, I expanded the bracket as:
(a+b x)^2=a^2+2abx+(bx)^2.

Then i evaluated the integral and simplifed the result to get:

$\frac{(b^2+2ab+a^2)c^2+(3b^2+8ab+5a^2)c+2b^2+6ab+6a^2}{c^3+6c^2+11c+6}=2$

and now i'm stuck.
Since we have linear and quadratic powers of b, i don't know how to make b the subject. any ideas will be very much appreciated.
Thank you

2. Jun 30, 2011

### micromass

What is the integral

$$\int_0^1{(a^2x^c+2abx^{c+1}+b^2x^{c+2})dx}$$

Don't "simplify" anything, just calculate the integral, nothing more.

3. Jun 30, 2011

### sara_87

after evaluating the integral, we have:

(a^2/(c+1)) + (2ab/(c+3))+(b^2/(c+3))=2

(this looks much cleaner

but still, how would i make b the subject?

thank you in advance

4. Jun 30, 2011

### micromass

Well, you have

$$\frac{1}{c+3}b^2+\frac{2a}{c+2}b+\frac{a^2-2c-2}{c+1}=0$$

this is a quadratic equation in b and can be solved by the quadratic formula.

5. Jun 30, 2011

### sara_87

so, i will have 2 solutions?

6. Jun 30, 2011

### micromass

Yes, unless the solutions coincide. But I doubt that will happen.

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