# Hi, I was wondering how transformations, such as dilations, affect definite integrals. Specifically, why is the first statement in the example below true? Why is the first integral 1/3 of the second? Thanks!

## The first statement is this ${\int}_{1}^{3} f \left(3 x + 1\right) \mathrm{dx} = \left(\frac{1}{3}\right) \cdot {\int}_{4}^{10} f \left(x\right) \mathrm{dx}$ I am given this information ${\int}_{4}^{10} f \left(x\right) \mathrm{dx} = 3$ Therefore ${\int}_{1}^{3} f \left(3 x + 1\right) \mathrm{dx} = \left(\frac{1}{3}\right) \cdot 3 = 1$ There is a dilation of 1/3 from the y-axis and a horizontal translation of 1/3 to the left, but I am not sure why the first integral is 1/3 of the second.

Sep 25, 2017

We aim to show that:

$I = {\int}_{1}^{3} f \left(3 x + 1\right) \mathrm{dx} = \left(\frac{1}{3}\right) \cdot 3 = 1$

Given:

${\int}_{4}^{10} f \left(x\right) \setminus \mathrm{dx} = 3$

We can perform a substitution on the first integral:

Let $u = 3 x + 1 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 3$

And when we perform a transformation substitution, we must change the limits of integration accordingly,

When $x = \left\{\begin{matrix}1 \\ 3\end{matrix}\right. \implies u = \left\{\begin{matrix}4 \\ 10\end{matrix}\right.$

Now, we can prepare and perform the substitution, and change the integration limits, which gives:

$I = {\int}_{1}^{3} \setminus f \left(3 x + 1\right) \setminus \mathrm{dx}$
$\setminus \setminus = {\int}_{1}^{3} \setminus f \left(3 x + 1\right) \setminus \left(\frac{1}{3}\right) \left(3\right) \mathrm{dx}$
$\setminus \setminus = {\int}_{4}^{10} \setminus f \left(u\right) \setminus \left(\frac{1}{3}\right) \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{3} \setminus {\int}_{4}^{10} \setminus f \left(u\right) \setminus \mathrm{du} \setminus \setminus \setminus \setminus$ ..... (A)
$\setminus \setminus = \frac{1}{3} \left(3\right) \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ ..... (B)
$\setminus \setminus = 1$ QED

Here we have used two fundamental properties of integrals, namely:

Scaling by a Constant

$\int \setminus c \setminus f \left(t\right) \setminus \mathrm{dt} = c \setminus \int \setminus f \left(t\right) \setminus \mathrm{dt}$ where $\setminus$ is constant

This was used at [A] above to factor out the $\frac{1}{3}$ multiplier.

Changing the variable alone of a definite Integration

${\int}_{a}^{b} \setminus f \left(x\right) \setminus \mathrm{dx} = {\int}_{a}^{b} \setminus f \left(z\right) \setminus \mathrm{dz} = {\int}_{a}^{b} \setminus f \left(t\right) \setminus \mathrm{dt}$

This was used at [B] above to evaluate:

${\int}_{4}^{10} \setminus f \left(u\right) \mathrm{du} \setminus \setminus \setminus$, which is the same as ${\int}_{4}^{10} \setminus f \left(x\right) \mathrm{dx}$

Not that here no transformation has occurred. If we think about a definite integral in term of the area under the curve, we are simply saying that if we change the label along on the graph of the function from$x$ to $t$, say the area does not change.

Further Explanation:

For further insight as to why the variables must be changed during a transformation, please review this solution: