The following three examples show how to find areas where the exponential function is involved.
Find the area enclosed between the curve $y=e^{-x}$y=e−x, 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 | $=$= | $\int_0^1e^{-x}dx$∫10e−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}$1−1e2, and for $x=3$x=3, $A=1-\frac{1}{e^3}$A=1−1e3. Clearly, for the line $x=n$x=n, we have $A=1-\frac{1}{e^n}$A=1−1en. 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.
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−∫lna0eydy |
$=$= | $a\ln a-\left[e^y\right]_0^{\ln a}$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 | |
Find the area between the curves $y=e^x$y=ex and $y=2-e^x$y=2−ex 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=2−ex 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 | $=$= | $\int_0^1e^xdx-\int_0^12-e^xdx$∫10exdx−∫102−exdx |
$=$= | $\int_0^1e^x-\left(2-e^x\right)dx$∫10ex−(2−ex)dx | |
$=$= | $2\int_0^1e^x-1dx$2∫10ex−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 |
Calculate the area enclosed between the curve $y=e^x$y=ex, the coordinate axes, and the line $x=2$x=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.
The functions $f\left(x\right)=3-e^x$f(x)=3−ex and $g\left(x\right)=e^x+1$g(x)=ex+1 intersect at point $A$A.
Solve for the $x$x-coordinate of $A$A. Write each line of working as an equation.
Find the area bound between the two curves and the line $x=\ln3$x=ln3.
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