1. ## integration problem

Given that $\int_0^a(x-1)dx=\frac{1}{4}\int_0^a(x+2)dx$,find the value of a.

2. Originally Posted by mastermin346
Given that $\int_0^a(x-1)dx=\frac{1}{4}\int_0^a(x+2)dx$,find the value of a.

Apply the Fundamental Theorem of Calculus to evaluate $\int_0^ax-1\,dx$ and $\tfrac{1}{4}\int_0^ax+2\,dx$.

At this point, $\int_0^ax-1\,dx=\tfrac{1}{4}\int_0^a x+2\,dx$ will be an equation in terms of 'a'; at this point, you can solve for 'a'.

Can you take it from here?

3. Originally Posted by Chris L T521
Apply the Fundamental Theorem of Calculus to evaluate $\int_0^ax-1\,dx$ and $\tfrac{1}{4}\int_0^ax+2\,dx$.

At this point, $\int_0^ax-1\,dx=\tfrac{1}{4}\int_0^a x+2\,dx$ will be an equation in terms of 'a'; at this point, you can solve for 'a'.

Can you take it from here?
hi,i have try many times,,but my answer not correct.

4. OK, just because you tried many times I will post the answer:
$\int_0^ax-1\,dx=\frac{1}{4}\int_0^ax+2\,dx$
$\mid(x^2-x)\mid_0^a=\frac{1}{4}\mid(x^2+2x)\mid_0^a$
$a^2-a=\frac{1}{4}(a^2+2a)$
After this, you should know how to solve by yourself.
My answer will be: $a=2$ because $a>0 \forall a\in R$

5. Originally Posted by hungthinh92
OK, just because you tried many times I will post the answer:
$\int_0^ax-1\,dx=\frac{1}{4}\int_0^ax+2\,dx$
$\mid({\color{red} x^2}-x)\mid_0^a=\frac{1}{4}\mid({\color{red} x^2}+2x)\mid_0^a$
$a^2-a=\frac{1}{4}(a^2+2a)$
After this, you should know how to solve by yourself.
My answer will be: $a=2$ because $a>0 \forall a\in R$

Its ${\color{white}x}\frac{1}{2}x^2$.

6. Its.
I did a really big mistake... Now the answer will be:
$\frac{1}{2}a^2-a=\frac{1}{4}(\frac{a^2}{2}+2a)$
And then $a = 4$ instead of 2

Thank General.