Solving an equation after integrating

[sara_87]

Homework Statement

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

Homework Equations

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

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

 

[micromass]

Homework Statement

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

Homework Equations

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

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

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.

 

[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

 

[micromass]

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

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.

 

[sara_87]

so, i will have 2 solutions?

 

[micromass]

so, i will have 2 solutions?

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