Question

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_1^3f(3x+1)dx=(1/3)*int_4^10f(x)dx`

I am given this information

`int_4^10f(x)dx=3`

Therefore

`int_1^3f(3x+1)dx=(1/3)*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.

Answer 1

We aim to show that:

`I = int_1^3f(3x+1)dx = (1/3)*3=1`

Given:

`int_4^10 f(x) \ dx = 3`

We can perform a substitution on the first integral:

Let `u=3x+1 => (du)/dx = 3`

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

When `x = { (1), (3) :} => u = { (4), (10) :}`

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

`I = int_1^3 \ f(3x+1) \ dx`
`\ \ = int_1^3 \ f(3x+1) \ (1/3)(3)dx`
`\ \ = int_4^10 \ f(u) \ (1/3) \ du`
`\ \ = 1/3 \ int_4^10 \ f(u) \ du \ \ \ \` ..... (A)
`\ \ = 1/3 (3) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \` ..... (B)
`\ \ = 1` QED

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

Scaling by a Constant

`int \ c \ f(t) \ dt = c \ int \ f(t) \ dt` where `\` is constant

This was used at [A] above to factor out the `1/3` multiplier.

Changing the variable alone of a definite Integration

`int_a^b \ f(x) \ dx = int_a^b \ f(z) \ dz = int_a^b \ f(t) \ dt`

This was used at [B] above to evaluate:

`int_4^10 \ f(u) du \ \ \`, which is the same as `int_4^10 \ f(x) 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:

Evaluating definite Integral and changing the values of A and B?