Proof by Mathematical Induction

We aim to prove that for all positive integers n :

A_n = n^2 \pi a^2

where A_n is the area enclosed between y = g'(x) , the x -axis, and the lines x = 0 and x = n a \pi .

Base Case: n = 1

For n = 1 :

The area A_1 is:

A_1=\int_0^{a\pi}|g^{\prime}(x)|\,dx

Since g'(x) = x \sin\left( \dfrac{x}{a} \right) and x > 0 , we need to consider the sign of \sin\left( \dfrac{x}{a} \right) on [0, a\pi] . On this interval, \sin\left( \dfrac{x}{a} \right) \geq 0 , so g'(x) \geq 0 . Thus:

A_1 = \int_{0}^{a \pi} g'(x) \, dx = g(a \pi) - g(0)

Given g(0) = 0 , we have:

A_1 = g(a \pi)

From part (d), the y -coordinate of the stationary point at x = a \pi is :

y_1 = a^2 (1) \pi (-1)^{1+1} = a^2 \pi

Therefore:

A_1 = y_1 = a^2 \pi = (1)^2 \pi a^2

The formula holds for n = 1 .

Inductive Step

Inductive Hypothesis:

Assume the formula holds for n = k , so:

A_k = k^2 \pi a^2

We need to prove: A_{k+1} = (k+1)^2 \pi a^2

Compute A_{k+1} :

The area A_{k+1} is:

A_{k+1} = \int_{0}^{(k+1) a \pi} | g'(x) | \, dx

This can be split into two parts:

From x = 0 to x = k a \pi : This area is A_k .
From x = k a \pi to x = (k+1) a \pi : Let's denote this area as \Delta A .

So:

A_{k+1} = A_k + \Delta A

Compute \Delta A :

The area \Delta A is:

\Delta A = \int_{k a \pi}^{(k+1) a \pi} | g'(x) | \, dx = | g\left( (k+1) a \pi \right) - g\left( k a \pi \right) |

This follows from the Fundamental Theorem of Calculus, considering the absolute value due to g'(x) changing sign.

Find g\left( (k+1) a \pi \right) and g\left( k a \pi \right) :

From part (d), the y -coordinate at x = n a \pi is:

y_{n}=a^2n\pi(-1)^{n+1}=\pm a^2n\pi

For n = k+1 :

\begin{align*}y_{k+1} & =a^2(k+1)\pi(-1)^{(k+1)+1}=\mp a^2\left(n+1\right)\pi\end{align*}

The signs are opposite as \left(-1\right)^{n+2}=-\left(-1\right)^{n+1}.

Compute \Delta A :

\begin{align*}\Delta A & =|y_{k+1}-y_{k}|\\ \placeholder{} & \placeholder{}\\ \placeholder{} & =a^2\pi\left|\mp\left(k+1\right)-\left(\pm k\right)\right|\\ \placeholder{} & \placeholder{}\\ \placeholder{} & =a^2\pi\left|\mp\left(k+1\right)\mp k\right|\\ \placeholder{} & \placeholder{}\\ \placeholder{} & =a^2\pi\left|\mp\left(2k+1\right)\right|\\ \placeholder{} & \placeholder{}\\ \placeholder{} & =\left(2k+1\right)a^2\pi\end{align*}

Compute A_{k+1} :

A_{k+1}=A_{k}+\Delta A

Since we assumed A_{k}=k^2\pi a^2 :

\begin{align}A_{k+1} & =k^2\pi a^2+\left(2k+1\right)\pi a^2\\ \placeholder{} & \placeholder{}\\ \placeholder{} & =\pi a^2\left(k^2+2k+1\right)\\ \placeholder{} & \placeholder{}\\ \placeholder{} & =\pi a^2\left(k+1\right)^2\end{align}

Hence the statement is true for n=k+1, if it is true for n=k.

Conclusion:

We have shown the statement is true for n=1, and that if it is true for n=k, then it is also true for n=k+1.

Hence, by the principle of mathematical induction, the formula:

A_n = n^2 \pi a^2

holds for all positive integers n .

Paper 3 Practice