NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Area bound by curves (exp)

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$10exdx
  $=$= $\left[-e^{-x}\right]_0^1$[ex]10
  $=$= $-\frac{1}{e}-\left(-1\right)$1e(1)
  $=$= $1-\frac{1}{e}$11e

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 $=$= $a\ln a-\int_0^{\ln a}e^ydy$alnalna0eydy
  $=$= $a\ln a-\left[e^y\right]_0^{\ln a}$alna[ey]lna0
  $=$= $a\ln a-\left(e^{\ln a}-e^0\right)$alna(elnae0)
  $=$= $a\ln a-(a-1)$alna(a1)
  $=$= $1-a+a\ln a$1a+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$10exdx102exdx
  $=$= $\int_0^1e^x-\left(2-e^x\right)dx$10ex(2ex)dx
  $=$= $2\int_0^1e^x-1dx$210ex1dx
  $=$= $2\left[e^x-x\right]_0^1$2[exx]10
  $=$= $2\left[e^x-x\right]_0^1$2[exx]10
  $=$= $2\left(e-2\right)\approx1.4366$2(e2)1.4366

More Worked Examples


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

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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.

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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.



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


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