Given that $\displaystyle \int_0^a(x-1)dx=\frac{1}{4}\int_0^a(x+2)dx$,find the value of a.
anyone help me please.
Apply the Fundamental Theorem of Calculus to evaluate $\displaystyle \int_0^ax-1\,dx$ and $\displaystyle \tfrac{1}{4}\int_0^ax+2\,dx$.
At this point, $\displaystyle \int_0^ax-1\,dx=\tfrac{1}{4}\int_0^a x+2\,dx$ will be an equation in terms of 'a'; at this point, you can solve for 'a'.
Can you take it from here?
OK, just because you tried many times I will post the answer:
$\displaystyle \int_0^ax-1\,dx=\frac{1}{4}\int_0^ax+2\,dx$
$\displaystyle \mid(x^2-x)\mid_0^a=\frac{1}{4}\mid(x^2+2x)\mid_0^a$
$\displaystyle a^2-a=\frac{1}{4}(a^2+2a)$
After this, you should know how to solve by yourself.
My answer will be: $\displaystyle a=2$ because $\displaystyle a>0 \forall a\in R$