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**.

## Expert Answer

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:

\[ = \dfrac{1}{2} \times 10 \]

## Numerical Answer

\[ \dfrac{1}{2} \times 10 = 5 \]

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

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,

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