Let $f$ be a differentiable function defined for all real $x$, where $f(x)\ge 0$ for all $x\in[0,a]$

Question

Let $f$ to be a differentiable function defined for all real $x$, where $f(x)\ge 0$ for all $x\in[0,a]$.If $$\int_0^a f(x)\,dx = a, $$ then $$2\int_0^{5a}\left(f\left( \frac x 5 \right) + 3 \right)\,dx$$ is equal to

Can anyone help me with this question. Don't know what to do.

Thanks.

Answer 1

By using the substitution $u=\frac{x}{5}$, you have : $$\begin{align}2\int_0^{5a}\left(f\left( \frac x 5 \right) + 3 \right)dx&=2\int_0^{a}(f (u) +3 )5du\\&=10\int_0^{a}(f(u)+3)du\\&=10\int_0^{a}f(u)du+10\int_0^{a}3du\\&=40a \end{align}$$

Answer 2

Firstly, split the integral $2\int_0^{5a}(f(\frac{x}{5})+3)dx=2\int_0^{5a}f(\frac{x}{5})+30a$. Let's define $u=\frac{x}{5}$ such that $2\int_0^{5a}f(\frac{x}{5})dx=10\int_0^{a}f(u)du=10a$, such that $2\int_0^{5a}(f(\frac{x}{5})+3)=10a+30a=40a$.

Answer 3

We can use substitution and properties of integrals to simplify this problem. We will get something easier to deal with: $2\int_0^{5a}(f(x/5)+3)dx$ = $2$ $[$ $\int_0^{5a}$ $f(x/5)$$dx$ + $\int_0^{5a}$$3$$dx$$ ].$ Let u = $x/5$. Then, $du = dx/5 $. So, we get: $2$ $[$ $\int_0^{5a}$ $f(u)$$5du$ + $\int_0^{5a}$$3$$dx$$ ].$ Can you finish it from here?