 # If f is continuous and integral 0 to 4 f(x)dx = 10 , find integral 0 to 2 f(2x)dx. This problem aims to find the integral of a continuous function given an integral of the same function at some other point. This problem requires the knowledge of basic integration along with the integration substitution method. A continuous function is a function with no disruption in the variation of the function, and this means that there is no abrupt change in the values, which is also called discontinuity.

The integral of any function is always continuous, but if that function is itself continuous, then its integral is differentiable.

Now, the problem states that:

if $\int_{0} ^ {4} f(x) \ ,dx$ $= 0$, then what $\int_{0} ^ {2} f(2x) \ , dx$ be equals to.

First, we will solve the integral $\int_{0} ^ {2} f(2x) \, dx$  by substituting $2x = u$. Now, let’s derivate it with respect to $x$, it gives us $2dx = du$, to write $dx$ in terms of $du$.

To eliminate x from the integral, we will multiply and divide $2$ to easily plug in the substitutions.

$= \dfrac{1}{2} \int_{0} ^ {2} f(2x) \, 2dx$

Since the independent variable has changed, its limits also need to be shifted.

So the limits will now change from $\int_{0 \times 2} ^ {2 \times 2}$ to $\int_{0} ^ {4}$.

Finally,

$= \dfrac{1}{2} \int_{0} ^ {4} f(u) \,du$

Remember, $\int_{a} ^ {b} f(x) \,dx = \int_{a} ^ {b} f(u) \,du$

We can rewrite our Integral as:

$= \dfrac{1}{2} \int_{0} ^ {4} f(x) \,dx$

As given in the statement, we can plug in the value $= \int_{0} ^ {4} f(x) \,dx = 10$.

Using this information, we can update the equation as:

$= \dfrac{1}{2} \times 10$

$\dfrac{1}{2} \times 10 = 5$

$\int_{0}^{2} f(2x) \,dx = 5$

This value is the area under the curve that represents the sum of infinite and indefinitely small quantities, just like when we multiply two numbers, one of them keeps producing different values.

## Example

If $f$ is continuous and integral $0$ to $4$ $f(x)dx = -18$ , find integral $0$ to $2$ $f(2x)dx$.

Substituting $2x = u$ and taking derivative, $2dx = du$.

Multiplying limits by $2$, we get:

$\int_{0 \times 2}^{2 \times 2} to \int_{0}^{4}$

Plugging in the substitutes, we get:

$= \dfrac{1}{2} \int_{0} ^ {4} f(u) \,du$

As we know, $\int_{a} ^ {b} f(x) \,dx = \int_{a} ^ {b} f(u) \, du$

Substituting the value of $\int_{0} ^ {4} f(x) \,dx = -18$

$= \dfrac{1}{2} \times -18$

$= -9$

Finally,

$\int_{0} ^ {2} f(2x) \,dx = -9$