What is the area under the curve y = cos x

What is the area under the curve $y = \cos x - \sin x +2$ ?

Question

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.

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.

Solution

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.

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).

What is the area under the curve

y = \cos x - \sin x +2

Question

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.

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.

Solution

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.

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

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).