Review question

# What is the area under the curve $y = \cos x - \sin x +2$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8679

## Solution

Given the curve $y = \cos x - \sin x +2$ find in terms of $a$ an expression for the area $A$ enclosed between the axes, the ordinate $a \; (> 0)$ and this curve. Verify that when $a = 1.5$, $A = 3.07$ approximately.

The ordinate $a$ means the line $x=a$. Our curve lies entirely above the $x$-axis. It is something like this.

To calculate the area, we integrate: \begin{align*} A &= \int_0^a \cos x- \sin x+2 \:dx \\ &= \big[\sin x + \cos x + 2x\big]_0^a \\ &= \sin a + \cos a + 2a -1. \end{align*}

Substituting $a = 1.5$ gives $A = 3.0682... \approx 3.07$ as required.

It is required to find the value of $a$ for which $A = 3$. Taking $1.5$ as a first approximation to $a$, find a second approximation, giving three significant figures in your answer.

So we need to find an approximate solution to $3 = \sin a + \cos a + 2a -1$ that is better than $a=1.5$.

There are a number of methods we could use here; decimal search, binary search, linear interpolation, the rearrangement method or simply zooming in with graphing software.

We will use the Newton-Raphson method. This says that given a guess $x_0$ at the root $\alpha$ (a solution to $f(x) = 0$), then as long as $x_0$ is reasonably close to $\alpha$, a better approximation to $\alpha$ is $x_1$ where

$x_1 = x_0 - \dfrac{f(x_0)}{f'(x_0)}.$

So our equation is $f(x) = 0$ where $f(x) = \sin x + \cos x + 2x - 4$, which gives $f'(x) = \cos x -\sin x + 2$.

Thus $x_1 = 1.5 - \dfrac{\sin 1.5 + \cos 1.5 + 3 - 4}{\cos 1.5 -\sin 1.5 + 2} = 1.4264...$

So we have a better approximation to $a$ than $1.5$, namely $1.43$ ($3$sf).

Check – $\sin 1.43 + \cos 1.43 + 2\times 1.43 -1 = 2.99$ ($3$sf).