# Area L

By sketching a graph of a continuous increasing function, can you prove a useful result about integrals?

## Problem

Let $f(x)$ be a continuous increasing function in the interval $a\leq x\leq b$ where $0 < a < b$ and $0\leq f(a) < f(b)$.

**Can you prove the following formula with the help of a sketch?**$$\int_{f(a)}^{f(b)} f^{-1}(t) \,dt + \int_a^bf(x) \,dx = bf(b) - af(a).$$**Why must $f(x)$ be increasing in the interval $a\leq x\leq b$?****How could you evaluate a similar integral if $f(x)$ is decreasing?**

Once you've proved the formula, find the value of $\int _1^4 \sqrt t \,dt$, in two different ways; firstly by evaluating the integral directly, and secondly by using the formula above with $f(x)=x^2$.

Have a go at using the formula to evaluate $\int_0^1\sin^{-1}t \,dt.$

Can you find other functions that you can integrate more easily using this formula than by other means?

## Getting Started

Start by drawing a sketch of the graph of $f(x)$ for $a\leq x \leq b$.

What is represented by $bf(b) - af(a)$?

Which parts of the graph correspond to the integrals?

## Student Solutions

We received solutions from Amrit, Adithya, Guruvignesh and Agathiyan, all from Hymers College, and Pablo from Kings College Alicante. Here is Pablo's solution:

As you can see in Graph I, the areas in purple and blue add up to the big rectangle, $bf(b)$, minus $af(a)$, the small rectangle.

There is a change of variable in $f^{-1}(t)$ because the area isn't taken

from the x-axis, but from the t-axis. $f^{-1}(x)$ is the inverse of $f(x)$. It is obtained either by switching variables and rearranging the equation again or by reflecting it in the line y=x.

The reason why $f(x)$ must be increasing in the interval $a \leq x \leq b$ is shown in Graph II. If it weren't, if at some point it reached a turning point and began to decrease, the area (orange) at the top would not be accounted for.

If $f(x)$ is always decreasing in the interval $a \leq x \leq b$, a similar formula comes out, but through more algebraic manipulation. The area called A in Graph III is:

$A = bf(a) - bf(b) - \int_{f(b)}^{f(a)} f^{-1}(t) d t$

but also, $A = bf(a) - af(a) \int_a^b f(x) d x$

so $$bf(a) - bf(b) - \int_{f(b)}^{f(a)} f^{-1}(t) d = bf(a) - af(a) \int_a^b f(x) d x$$ $$\int_a^b f(x) d x = bf(a) - af(a) - \left[bf(a) - bf(b) - \int_{f(b)}^{f(a)} f^{-1}(t) d t\right]$$ $$\int_a^b f(x) d x = bf(b) - af(a) + \int_{f(b)}^{f(a)} f^{-1}(t) d t $$

The difference with the first formula is that $\int_{f(b)}^{f(a)} f^{-1}(t) d t$ is added, not subtracted. Additionally, the boundaries are swapped, as $f(b)$ is less than $f(a)$.

First, we can find the value of
$\int_1^4\sqrt t dt$ the conventional

way:

$$\eqalign{\int_1^4 \sqrt x dx &= \left[\frac23 x^\frac32\right]_1^4 \\

&= \frac23 ( 4^\frac32 - 1^\frac32) \\

&= \frac{14}3}$$

Now using our formula:

$$\eqalign{\int_1^4 \sqrt x dx &= bf(b) - af(a) - \int_{f(a)}^{f(b)}f^{-1}(t)dt \\

&= 4\sqrt4 - 1\sqrt1 - \int_{\sqrt(1)}^{\sqrt(4)} t^2 dt \\

&= 8 - 1 - \left[\frac13 t^3 \right]_1^2 \\

&= 7 - \frac83 + \frac13 \\

&= \frac{14}3}$$

Which gives the same answer!

For $\sin^{-1}(x)$, using Graph V:

$$ \eqalign{ \int_0^1 \sin^{-1}(x) dx &= bf(b) - af(a) - \int_{f(a)}^{f(b)}f^{-1}(t)dt \\ &= 1\times \sin^{-1}(1) - 0\times \sin^{-1}(0) - \int_{f(a)}^{f(b)} \sin(t) dt \\ &= \frac{\pi}2 - 0 - \int_0^{\frac{\pi}2} \sin(t) dt \\ &=\frac{\pi}2 - \left[-\cos(t)\right]_0^{\frac{\pi}2} \\ &= \frac{\pi}2 - \left[(-cos(\frac{\pi}2)-(-cos(0))\right]\\ &= \frac{\pi}2 - 1}$$

Another function that can be integrated using this method is $ln(x)$:

$$ \eqalign{\int_1^{e^2} ln(x) dx &= e^2ln(e^2) - 1ln(1) - \int_0^2 e^t dt \\ &= 2e^2 - 0 - \left[e^t\right]_0^2 \\ &= 2e^2 - (e^2 - e^0) \\

&= 2e^2 - e^2 + 1 \\ &= e^2 + 1}$$

Image

As you can see in Graph I, the areas in purple and blue add up to the big rectangle, $bf(b)$, minus $af(a)$, the small rectangle.

There is a change of variable in $f^{-1}(t)$ because the area isn't taken

from the x-axis, but from the t-axis. $f^{-1}(x)$ is the inverse of $f(x)$. It is obtained either by switching variables and rearranging the equation again or by reflecting it in the line y=x.

Image

The reason why $f(x)$ must be increasing in the interval $a \leq x \leq b$ is shown in Graph II. If it weren't, if at some point it reached a turning point and began to decrease, the area (orange) at the top would not be accounted for.

Image

If $f(x)$ is always decreasing in the interval $a \leq x \leq b$, a similar formula comes out, but through more algebraic manipulation. The area called A in Graph III is:

$A = bf(a) - bf(b) - \int_{f(b)}^{f(a)} f^{-1}(t) d t$

but also, $A = bf(a) - af(a) \int_a^b f(x) d x$

so $$bf(a) - bf(b) - \int_{f(b)}^{f(a)} f^{-1}(t) d = bf(a) - af(a) \int_a^b f(x) d x$$ $$\int_a^b f(x) d x = bf(a) - af(a) - \left[bf(a) - bf(b) - \int_{f(b)}^{f(a)} f^{-1}(t) d t\right]$$ $$\int_a^b f(x) d x = bf(b) - af(a) + \int_{f(b)}^{f(a)} f^{-1}(t) d t $$

The difference with the first formula is that $\int_{f(b)}^{f(a)} f^{-1}(t) d t$ is added, not subtracted. Additionally, the boundaries are swapped, as $f(b)$ is less than $f(a)$.

First, we can find the value of

Image

way:

$$\eqalign{\int_1^4 \sqrt x dx &= \left[\frac23 x^\frac32\right]_1^4 \\

&= \frac23 ( 4^\frac32 - 1^\frac32) \\

&= \frac{14}3}$$

Now using our formula:

$$\eqalign{\int_1^4 \sqrt x dx &= bf(b) - af(a) - \int_{f(a)}^{f(b)}f^{-1}(t)dt \\

&= 4\sqrt4 - 1\sqrt1 - \int_{\sqrt(1)}^{\sqrt(4)} t^2 dt \\

&= 8 - 1 - \left[\frac13 t^3 \right]_1^2 \\

&= 7 - \frac83 + \frac13 \\

&= \frac{14}3}$$

Which gives the same answer!

For $\sin^{-1}(x)$, using Graph V:

Image

$$ \eqalign{ \int_0^1 \sin^{-1}(x) dx &= bf(b) - af(a) - \int_{f(a)}^{f(b)}f^{-1}(t)dt \\ &= 1\times \sin^{-1}(1) - 0\times \sin^{-1}(0) - \int_{f(a)}^{f(b)} \sin(t) dt \\ &= \frac{\pi}2 - 0 - \int_0^{\frac{\pi}2} \sin(t) dt \\ &=\frac{\pi}2 - \left[-\cos(t)\right]_0^{\frac{\pi}2} \\ &= \frac{\pi}2 - \left[(-cos(\frac{\pi}2)-(-cos(0))\right]\\ &= \frac{\pi}2 - 1}$$

Image

Another function that can be integrated using this method is $ln(x)$:

$$ \eqalign{\int_1^{e^2} ln(x) dx &= e^2ln(e^2) - 1ln(1) - \int_0^2 e^t dt \\ &= 2e^2 - 0 - \left[e^t\right]_0^2 \\ &= 2e^2 - (e^2 - e^0) \\

&= 2e^2 - e^2 + 1 \\ &= e^2 + 1}$$

## Teachers' Resources

### Why do this problem?

This problem looks more complicated than it is. Students need to interpret the notation and use their understanding of definite integrals as areas under curves to prove a formula graphically. Once they make sense of the formula in terms of a diagram, they can use it to evaluate integrals which might prove
challenging to calculate by other means.

### Possible approach

Students could sketch a suitable continuous increasing function on paper or whiteboards, and mark on suitable points $a, b, f(a)$ and $f(b)$. Give them time in groups to identify the areas on their graph represented by $\int_{f(a)}^{f(b)} f^{-1}(t) \,dt$, $\int_a^bf(x) \,dx$, $bf(b)$ and $af(a)$.Once they are convinced that the proof of the formula follows from a diagram, give them the chance to verify the formula for cases they can integrate directly, like $\int_a^b \sqrt{t} \,dt$.

The problem suggests using the formula to evaluate $\int_0^1 \sin^{-1}t \,dt$ but other inverse functions of continuous increasing functions can be used, and it offers good integration practice to allow students to suggest functions of their own to integrate which they can evaluate by other means as a check. It is worth taking the time to discuss the requirement for the function to be increasing between $a$ and $b$, and setting the challenge to find a similar formula for a function which is decreasing.

Key questions

On the graph $y=f(x)$, what is represented by $\int_{f(a)}^{f(b)} f^{-1}(t) \,dt$?

Why does the formula specify that $f(x)$ is increasing?

What could be done if $f(x)$ were decreasing?