Question

If $f$ is continuous and $\int\limits_0^4 {f(x)\,dx} = - 18$, how to evaluate $\int\limits_0^2 {f(2x)\,dx}$?

Answer 1

Here in this question they have mentioned the given function will be a continuous function and they have not mentioned the function. The definite integral is applied to the function and the final answer is obtained for the $\int\limits_0^4 {f(x)\,dx}$, by considering this and using the substitution property we are going to determine the value of $\int\limits_0^2 {f(2x)\,dx}$.

Complete step by step solution:
In integration we have two kinds of integrals, one is definite integral and another one is indefinite integral. Here the give question is related to the definite integral.
Now in the question they have given $\int\limits_0^4 {f(x)\,dx} = - 18$
Here we have to find the $\int\limits_0^2 {f(2x)\,dx}$
Now we substitute $2x = u$, on differentiating this function we get $2dx = du$, therefore we have $dx = \dfrac{{du}}{2}$. Now we will substitute the value of $2x$ and $dx$ in $\int\limits_0^2 {f(2x)\,dx}$. here the integral value will change.
here we have $2x = u$, when $x = 0$, the value of $u = 0$and when $x = 2$, the value of $u = 4$
Therefore $\int\limits_0^2 {f(2x)\,dx}$ can be written as
$\Rightarrow \int\limits_0^4 {f(u)\,\dfrac{{du}}{2}}$
So we take out $\dfrac{1}{2}$ from the integral and it is written as
$\Rightarrow \dfrac{1}{2}\int\limits_0^4 {f(u)\,du}$
We know that $\int\limits_0^4 {f(x)\,dx} = - 18$ , so on considering this and above function is written as
$\Rightarrow \dfrac{1}{2} \times - 18$
On simplifying we have
$\Rightarrow - 9$
Therefore, the value of $\int\limits_0^2 {f(2x)\,dx} = - 9$.

Note:
The integration is an inverse of the differentiation. To use integration we have a standard formula for some terms. Here the question is a general one. We use the substitution method and we write the given function and hence we use the simple arithmetic operations and solve the given function.