 New Zealand
Level 8 - NCEA Level 3

Area bound by curves (exp)

Lesson

The following three examples show how to find areas where the exponential function is involved.

Example 1

Find the area enclosed between the curve $y=e^{-x}$y=ex, the axes and the line $x=a$x=a where $a=1,2,3,...$a=1,2,3,...etc.

What can you conclude about the open area defined by the curve and the positive sides of the axes?

Choosing $a=1$a=1, we have the area $A$A determined as:

 $A$A $=$= $\int_0^1e^{-x}dx$∫10​e−xdx $=$= $\left[-e^{-x}\right]_0^1$[−e−x]10​ $=$= $-\frac{1}{e}-\left(-1\right)$−1e​−(−1) $=$= $1-\frac{1}{e}$1−1e​ For the line $x=2$x=2, we see that the area would become $1-\frac{1}{e^2}$11e2, and for $x=3$x=3, $A=1-\frac{1}{e^3}$A=11e3. Clearly, for the line $x=n$x=n, we have $A=1-\frac{1}{e^n}$A=11en. Since the positive fraction $\frac{1}{e^n}$1en becomes smaller and smaller as $n$n gets larger and larger, the area moves closer and closer to $1$1. So, even though the area is not completely enclosed by lines and/or curves, it will never exceed $1$1.

Example 2

The function $y=\ln x$y=lnx and $y=e^x$y=ex are inverses of each other. Show how we can find the area under the curve $y=\ln x$y=lnx between $x=1$x=1 and $x=a$x=a by considering the area bounded by the curve $y=e^x$y=ex, the $y$y axis and the line $y=\ln a$y=lna. The equation $y=\ln x$y=lnx can be rearranged to $x=e^y$x=ey, so we can consider the curve as a function of $y$y with $y$y as the independent variable. Thus we could write $f\left(y\right)=e^y$f(y)=ey

Hence, a way to find the area shown as $A_1$A1 in the diagram is to find the area of the rectangle defined by the the sum $A_1+A_2$A1+A2 and then subtract the area $A_2$A2 defined by  $\int_0^{\ln a}f(y)dy=\int_0^{\ln a}e^ydy$lna0f(y)dy=lna0eydy.

Thus we have:

 $A_1$A1​ $=$= $a\ln a-\int_0^{\ln a}e^ydy$alna−∫lna0​eydy $=$= $a\ln a-\left[e^y\right]_0^{\ln a}$alna−[ey]lna0​ $=$= $a\ln a-\left(e^{\ln a}-e^0\right)$alna−(elna−e0) $=$= $a\ln a-(a-1)$alna−(a−1) $=$= $1-a+a\ln a$1−a+alna
Example 3

Find the area between the curves $y=e^x$y=ex and $y=2-e^x$y=2ex and the line $x=1$x=1.

A sketch is a good place to start. Note that curve given by $y=2-e^x$y=2ex can be thought of as the curve given by $y=e^x$y=ex reflected in the $x$x axis and vertically translated up by $2$2 units. The area depicted in green is found as follows:

 $A$A $=$= $\int_0^1e^xdx-\int_0^12-e^xdx$∫10​exdx−∫10​2−exdx $=$= $\int_0^1e^x-\left(2-e^x\right)dx$∫10​ex−(2−ex)dx $=$= $2\int_0^1e^x-1dx$2∫10​ex−1dx $=$= $2\left[e^x-x\right]_0^1$2[ex−x]10​ $=$= $2\left[e^x-x\right]_0^1$2[ex−x]10​ $=$= $2\left(e-2\right)\approx1.4366$2(e−2)≈1.4366

More Worked Examples

QUESTION 1

Calculate the area enclosed between the curve $y=e^x$y=ex, the coordinate axes, and the line $x=2$x=2.

QUESTION 2

Consider the curve $y=\ln x$y=lnx which has been graphed below.

Solve for the exact area bounded by the curve, the coordinate axes, and the line $y=2$y=2.

QUESTION 3

The functions $f\left(x\right)=3-e^x$f(x)=3ex and $g\left(x\right)=e^x+1$g(x)=ex+1 intersect at point $A$A.

1. Solve for the $x$x-coordinate of $A$A. Write each line of working as an equation.

2. Find the area bound between the two curves and the line $x=\ln3$x=ln3.

Outcomes

M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

91579

Apply integration methods in solving problems