Applications of Integration

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The syllabus expects you to know all of the following applications for integration. Can you do them without even thinking?

Areas under curves
Let's start with the graph of \(y=x-x^3\). We'll ignore stationary points for now. (Can you reproduce it?)

We wish to find the area of the region A in the diagram. To do this, we first need to figure out the interval our area is defined over. Here, they are determined by the two right-most \(x\)-intercepts, i.e. \(x=0\) and \(x=1\). (Can you figure out why?)

Then, \( \int_a^b f(x)\,dx \) immediately gives us the area under the curve. So here, all we need to do is compute:
\begin{align*}
A &= \int_0^1 x - x^3\,dx\\
&= \left[ \frac{x^2}{2} - \frac{x^4}{4}\right]_0^1\\
&= \frac14 - 0\\
&= \frac14 \text{ units}^2.
\end{align*}
However, remember that the integral gives a signed-area quantity. Let's say instead of region A, I wanted the region 'opposite it' - the one below the \(x\)-axis, between \(x=-1\) and \(x=0\). If I computed \( \int_{-1}^0 x-x^3\,dx \), I would arrive at \(-\frac14\) - a negative answer! The idea is that if I want only the magnitude of the area, I would need to take absolute values on the final answer.

Total change given instantaneous rate of change
An average rate of change simply averages out how much a quantity changed over a fixed period of time. An instantaneous/marginal rate of change gives us precisely how much the quantity increases/decreases at a certain point in time. For example, how much water flows out of a tap.

If we know this instantaneous rate of change, we can always determine how much the original quantity changed over time. For example, suppose that a radioactive substance decays with a rate of \( \dfrac{dM}{dt} = -500 e^{-0.5t} \), where \(M\) is its mass measured in kilograms. Then, we can integrate to arrive at
\[ M = \int -500e^{-0.5t}\,dt = 250e^{-0.5t} + c. \]
For example, suppose I want how much of the quantity decayed between times \(t=5\) and \(t=20\). Then I can take the differences of the values for when I sub \(t=5\) and \(t=20\) in for this total change!
\[ \Delta M = (250e^{-2.5t} + c) - (250e^{-10t}+c) \approx 20.51\text{ kg}.\]
If I'm told some condition, I can also find what my value of \(c\) should be. For example, if I state that the initial mass was \(300\) kg, I can sub \(t=0\) and \(M=300\) in to obtain:
\[ 300 = 250e^0 + c \implies c = 50 .\]
This would then allow me to find the mass at any point in time, by computing \( M= 250e^{-0.5t}+50\).

Areas between curves
Of further interest is when the area isn't exclusively under one curve, but bound between two curves. Let's say we wanted the area between \(y=x^2\) and \(y=4-x^2\), in the first quadrant. The region has been tagged as B.

At times like these, when we're not told our boundaries of integration, we (again) have to find them. Here, the intention is for our lower boundary to be at \(x=0\), but our upper boundary is determined by the (right-most) point of intersection of the curves! We require simultaneous equations to track this down.
\begin{align*}
x^2&=4-x^2\\
2x^2&=4\\
x^2&=2\\
x&=\pm \sqrt{2}
\end{align*}
Since we want the point in the first quadrant, our \(x\)-coordinate is positive, so we take \(x=\sqrt{2}\). Our boundaries are now \( 0\) and \(\sqrt{2}\).

Then, to find the area, we always consider \( \boxed{\text{upper curve's equation} - \text{lower curve's equation}} \). No need for absolute values or any bizarre stuff here - just remember that rule!

However, note that the upper and lower curves are defined exclusively on the region we wish to integrate over. Whilst it is true that \(y=x^2\) is the upper curve for, say, \(x > 2\), it is in fact the lower curve for \(0 < x < 2\). So we need to compute:
\begin{align*}
A&= \int_0^2 (4-x^2) - x^2\,dx\\
&= \int_0^2 4-2x^2\,dx\\
&= \left[4x - \frac{2x^3}{3} \right]_0^2\\
&= \left(8-\frac{16}{3} \right) - (0-0)\\
&= \frac{8}{3}\text{ units}^2
\end{align*}

Retrieving displacement from acceleration
Recall that the acceleration of the particle is just the second derivative of its displacement. Furthermore, the velocity is the first derivative of the displacement.
\begin{align*}
s^{\prime\prime}(t) &= a(t)\\
s^\prime(t) = v(t)
\end{align*}
So if we're told the acceleration, we should expect that we can integrate our way back up to the displacement!

For example, let's take a particle with acceleration \(a(t) = 1 + e^{-t}\). We're also told that \(v(0) = 0\), and \(s(0) = -10\). Then using integration:
\begin{align*}
v(t) &= \int 1-e^{-t}\,dt\\
&= t+e^{-t} +c_1
\end{align*}
Using the fact that \(v(0) = 0\) we obtain \(0 = 0 + 1 + c_1\), so \( \boxed{c_1 = -1} \). This now gives us a nice expression for \(v(t)\), which we can use to find \(s(t)\):
\begin{align*}
v(t) &= t+e^{-t} - 1\\
s(t) &=\int t+e^{-t} - 1\,dt\\
&= \frac{t^2}{2} - e^{-t} -t + c_2.
\end{align*}
Now using \(s(0) = -10\), we obtain \( -10 = 0 - 1 - 0 + c_2 \), so \( \boxed{c_2 = -9} \). Thus
\[ s(t) = \frac{t^2}{2} - t - 9 - e^{-t} \]