You may also like

problem icon

Integral Equation

Solve this integral equation.

problem icon

Integral Sandwich

Generalise this inequality involving integrals.

problem icon

Integral Inequality

An inequality involving integrals of squares of functions.

Area L

Age 16 to 18 Challenge Level:
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
$$\eqalign{\int_1^4 \sqrt x dx &= \left[\frac23 x^\frac32\right]_1^4 \\
                                 &= \frac23 ( 4^\frac32 - 1^\frac32) \\
                                 &= \frac{14}3}$$

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