Applications of Integration (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Area below the x-axis

Consider the function $y = e^x - e$. We will find the area between this curve, the $x$-axis, and the $y$-axis.

Step 1: Sketch the curve

Start with the basic exponential curve $y = e^x$ and shift it down by $e$ units to obtain $y = e^x - e$.

Step 2: Find the intercepts

To find the $y$-intercept, substitute $x = 0$:

$y = e^0 - e$

$y = 1 - e$

To find the $x$-intercept, set $y = 0$:

$0 = e^x - e$

$e^x = e$

$x = 1$

Step 3: Identify the asymptote

The horizontal asymptote of $y = e^x$ is $y = 0$. After shifting down by $e$ units, the new horizontal asymptote is $y = -e$.

Step 4: Calculate the area

The region we want lies between $x = 0$ and $x = 1$, below the $x$-axis.

$\int_0^1 (e^x - e) \, dx = [e^x - ex]_0^1$

Note that $e$ is a constant, so $\int e \, dx = ex$.

$= (e^1 - e \cdot 1) - (e^0 - e \cdot 0)$

$= (e - e) - (1 - 0)$

$= 0 - 1$

$= -1$

The integral is negative because the region lies below the x-axis. Therefore, the required area is 1 square unit.

Worked Example: Area between exponential and polynomial curves

Find the area between the curves $y = e^x$ and $y = x^2$ from $x = 0$ to $x = 2$.

Step 1: Sketch both curves

For $x > 0$, the exponential curve $y = e^x$ is always above the parabola $y = x^2$. This is important because it determines which function to subtract from which.

Step 2: Apply the area formula

Since $y = e^x$ is above $y = x^2$ throughout the interval $0 \le x \le 2$, we integrate:

$\text{Area} = \int_0^2 (e^x - x^2) \, dx$

Step 3: Integrate each term

$= \left[e^x - \frac{1}{3}x^3\right]_0^2$

Step 4: Evaluate at the limits

$= \left(e^2 - \frac{8}{3}\right) - (e^0 - 0)$

$= \left(e^2 - \frac{8}{3}\right) - 1$

$= e^2 - \frac{8}{3} - 1$

$= e^2 - \frac{8}{3} - \frac{3}{3}$

$= e^2 - \frac{11}{3}$

$= e^2 - 3\frac{2}{3} \text{ square units}$