November 2, 2024 · 11:45 am
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Effect of Function Transformations on Integration

My year 12 Mathematical Methods students have questions like this

Given that f(x) is continuous everywhere and that \int_{4}^{10} f(x) dx=-10, find:

(a) \int_{4}^{10}2x +f(x) dx

(b) \int_{5}^{11} f(x-1) dx

(c) \int_{1}^{3} f(3x+1) dx

(d) \int_{-10}^{-4} -f(-x) dx

(e) \int{10}^{22} f(\frac{x-2}{2}) dx

(f) \int_{-3}^{-9} f(1-x) dx

OT Lee Mathematics Methods Textbook Ex 8.3 question 6

For the most part these questions aren’t too difficult, but the horizontal dilations cause issues.

(a) \int_{4}^{10} 2x +f(x) dx
\int_{4}^{10} 2x dx +\int_{2}^{10} f(x) dx
(x^2]_4^{10} + (-10)
10^2-4^2-10
=74

(b) \int_{5}^{11} f(x-1) dx
=-10




(c) \int_{1}^{3} f(3x+1) dx
Let u=3x+1
\frac{du}{dx}=3
dx=\frac{du}{3}

When x=1, u=4 and when x=3, u=10
\int_{4}^10 f(u) \frac{du}{3}
=\frac{1}{3}\times (-10)
=\frac{-10}{3}

(d) \int_{-10}^{-4} -f(-x) dx
-\int_{-10}^{-4} f(-x) dx

Let u=-x
\frac{du}{dx}=-1
dx=-du

When x=-4, u=4 and when x=-10, u=10
-\int_{4}^{10} f(u) -dx
=-10

(e) \int_{10}^{22} f(\frac{x-2}{2} dx
Let u=f(\frac{x-2}{2})
\frac{du}{dx}=\frac{1}{2}
\du=2dx

When x=10, u=4 and when x=22, u=10
2\int_{4}^{10} f(u) du
=-20

(f) \int_{-3}^{-9} 2f(1-x) dx
Let u=1-x
\frac{du}{dx}=-1

When x=-3, u=4 and when x=-9, u=10
=-2\int_{4}^{10} f(u) du
=20


Split the integral
Integrate the first part.


This is a horizontal translation (one unit to the right) so the shape of the curve doesn’t change.
The integration bounds have also shifted one unit to the right.




This is a horizontal dilation and translation. The easiest method is to use a change of variable





































Once you get the hang of it, you can skip the change of variable and multiply the value of the definite integral by the scale factor of the horizontal dilation (only if the integration bounds are also changed).

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