Category Archives: Year 12 Mathematical Methods

November 2, 2024 · 11:45 am

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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Filed under Definite, Integration, Uncategorized, Year 12 Mathematical Methods

September 24, 2024 · 3:30 am

Optimisation

An optimisation question from the 2019 ATAR Mathematics Methods exam.

I always like optimisation questions. There is a nice process to follow:

  • Find the function to optimise (in terms of one variable).
  • Find the stationary points.
  • Find the nature of the stationary points.
  • Find the maximum or minimum.
(a) Volume of the cylinder V=\pi r^2h
42=2r+h
h=42-2r
\therefore V_C=\pi r^2(42-2r)
Volume of spherical decorations V_S=\frac{4}{3}\pi( r_s)^3 where r_s=\frac{r}{3}
V_S=\frac{4\pi r^3}{81}
Volume unused space V=\pi r^2(42-2r)-20(\frac{4\pi r^3}{81})
V=2\pi (21r^2-r^3-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{81r^3}{81}-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{121r^3}{81})

(b) V=2\pi (21r^2-\frac{121r^3}{81})
\frac{dV}{dr}=2\pi (42r-\frac{121r^2}{27})
\frac{dV}{dr}=0
0=42r-\frac{121r^2}{27}
0=r(42-\frac{121r}{27})
r=0 or r=\frac{1134}{121}=9.372

\frac{d^2V}{dr^2}=2\pi (42-\frac{242r}{27})
(\frac{d^2V}{dr^2})_{|r=9.372}=-42
\therefore r=9.372 is a maximum.

Dimensions of the vase, internal diameter=18.7cm internal height=23.3cm

(c) Maximum volume of empty space =2\pi (21r^2-\frac{121r^3}{81})=3863.08cm^3
Volume of one sphere =\frac{4}{3}\pi r^3=3448.03cm^3

There is enough unused space for one extra decoration, but it would depend on how they are packed.

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Filed under Differentiation, Optimisation, Year 12 Mathematical Methods

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