Consider the following examples carefully. The areas shown in each graph involves evaluating integrals of functions that lead directly to log functions.
Remember that areas are absolute quantities - they are evaluated as positive numbers. Sometimes an integration evaluation needs to be interpreted to arrive at the correct result.
Find the area under the curve $y=\frac{16}{x-1}$y=16x−1 between $x=3$x=3 and $x=9$x=9 .
The sketch is quite straight forward - a hyperbola translated $1$1 unit to the right with a dilation factor of $16$16. We could find some points on the by thinking about certain values of $x$x. For example $x=5$x=5 means that $y=\frac{16}{5-1}=4$y=165−1=4. At $x=9$x=9, $y=\frac{16}{9-1}=2$y=169−1=2 etc. The vertical asymptote is given by $x=1$x=1.
The area is simply given by:
$A$A | $=$= | $\int_3^9\frac{16}{x-1}dx$∫9316x−1dx |
$=$= | $16\left[\ln\left(x-1\right)\right]_3^9$16[ln(x−1)]93 | |
$=$= | $16\left(\ln8-\ln2\right)$16(ln8−ln2) | |
$=$= | $16\left(3\ln2-\ln2\right)$16(3ln2−ln2) | |
$=$= | $32\ln2\approx22.18$32ln2≈22.18 | |
Hence the area is approximately $22.18$22.18 $u^2$u2.
Find the area under the curve $y=\frac{3}{6-4x}$y=36−4x between $x=0$x=0 and $x=1$x=1
The inverted hyperbola shown below has its vertical asymptote at $x=1.5$x=1.5 and a dilation factor of $3$3.
The area sought is found as follows:
$A$A | $=$= | $\int_0^1\frac{3}{6-4x}dx$∫1036−4xdx |
$=$= | $-\frac{3}{4}\int_0^1\frac{-4}{6-4x}dx$−34∫10−46−4xdx | |
$=$= | $-\frac{3}{4}\left[\ln\left(6-4x\right)\right]_0^1$−34[ln(6−4x)]10 | |
$=$= | $-\frac{3}{4}\left(\ln2-\ln6\right)$−34(ln2−ln6) | |
$=$= | $\frac{3}{4}\left(\ln6-\ln2\right)\approx0.824$34(ln6−ln2)≈0.824 | |
Find the area enclosed between the parabola $y=x^2$y=x2 and the hyperbola $y=6-\frac{6}{x+1}$y=6−6x+1 in the first quadrant.
Firstly, we need to find exactly where these two curves intersect each other, and to this end we equate both functions and solve for $x$x as follows:
$x^2$x2 | $=$= | $6-\frac{6}{x+1}$6−6x+1 |
$x^2\left(x+1\right)$x2(x+1) | $=$= | $6\left(x+1\right)-6$6(x+1)−6 |
$x^3+x^2$x3+x2 | $=$= | $6x$6x |
$x\left(x^2+x-6\right)$x(x2+x−6) | $=$= | $0$0 |
$x\left(x+3\right)\left(x-2\right)$x(x+3)(x−2) | $=$= | $0$0 |
$\therefore$∴ $x$x | $=$= | $0,-3,2$0,−3,2 |
The two intersections that interest us are $x=0$x=0 and $x=2$x=2, and in this respect the the two given curves intersect at the origin and at the point $\left(2,4\right)$(2,4). Here is the graph.
The area of interest, shaded in green, is evaluated as follows:
$A$A | $=$= | $\int_0^2\left(6-\frac{6}{x+1}\right)dx-\int_0^2x^2dx$∫20(6−6x+1)dx−∫20x2dx |
$=$= | $\int_0^2\left(6-\frac{6}{x+1}-x^2\right)dx$∫20(6−6x+1−x2)dx | |
$=$= | $\left[6x-6\ln\left(x+1\right)-\frac{x^3}{3}\right]_0^2$[6x−6ln(x+1)−x33]20 | |
$=$= | $\left(12-6\ln3-\frac{8}{3}\right)-\left(0\right)\approx2.74166$(12−6ln3−83)−(0)≈2.74166 | |
Hence, the area enclosed is approximately $2.742$2.742 $u^2$u2.
The curve given by $y=\frac{12x^2}{x^3-1}$y=12x2x3−1 is shown below.
It has discontinuity at $x=1$x=1, and a double root (implying a turning point) at the origin.
Using calculus techniques, we could confirm the existence of another stationary point, but its exact location requires solving a difficult cubic equation.
A modern computer program provides us with a way through. It reveals a minimum turning point near $\left(-1.26,-6.35\right)$(−1.26,−6.35).
Based on this, Richard has estimated the size of the enclosed area between the curve, the line $x=-4$x=−4, and the $x$x axis as being approximately $16u^2$16u2. Use calculus to find the exact area.
Looking at the figure, Richard's guess seems reasonable. We can imagine the green area being reshaped into a $4$4 by $4$4 square fairly snugly.
Before we proceed, we need to think about two things:
Firstly notice from the graph that an integral evaluated between $x=-4$x=−4 and $x=0$x=0 will provide a negative answer, and so we will need to change the sign of our answer to determine the area.
Secondly, in the evaluation of the integral, we will need to use absolute value signs around the argument of the derived logarithm, in order to keep it properly defined.
We proceed as follows:
$\int_{-4}^0\frac{12x^2}{x^3-1}dx$∫0−412x2x3−1dx | $=$= | $4\int_{-4}^0\frac{3x^2}{x^3-1}dx$4∫0−43x2x3−1dx |
$=$= | $4\left[\ln\left|x^3-1\right|\right]_{-4}^0$4[ln|x3−1|]0−4 | |
$=$= | $4\left(\ln1-\ln65\right)\approx-16.6975$4(ln1−ln65)≈−16.6975 | |
Hence, the area of the region is approximately $16.6975$16.6975 $u^2$u2.
Consider the function $y=\frac{2}{3x+5}$y=23x+5.
Sketch a graph of the function.
Find the area enclosed by the curve, the coordinate axes and the line $x=4$x=4.
Find the area enclosed by the function $f\left(x\right)=\frac{x}{x^2+2}$f(x)=xx2+2, the $x$x-axis and the lines $x=2$x=2 and $x=4$x=4.
Consider the function $y=x+\frac{1}{x}$y=x+1x.
State the domain of the function, using interval notation and commas where necessary.
What function value does the function approach as $x$x approaches $\infty$∞?
$y=\frac{1}{x}$y=1x
$y=0$y=0
$y=x$y=x
$y=\frac{1}{x}$y=1x
$y=0$y=0
$y=x$y=x
Find the area enclosed by the function, the $x$x-axis, and the lines $x=2$x=2 and $x=12$x=12.
Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods
Apply integration methods in solving problems