If a curve or a hybrid function looks like a familiar geometric shape when we graph it, we are easily able to find the area bounded by the curve(s) and the $x$x-axis.
To find the shaded area on this graph we can simply calculate the area of a right triangle.
Area=$\frac{3\times6}{2}$3×62
Area=$9$9 $units^2$units2
To find the shaded area bounded by this curve and the x-axis we can find the area of a semicircle.
Area=$\frac{\pi\times4^2}{2}$π×422
Area=$25.13$25.13 $units^2$units2
How would we find the area bounded by the three lines and the $x$x-axis below?
We don't need to break it up into three different sections, although we could. Instead we can find the area of a trapezium.
Area=$\frac{2\left(2+5\right)}{2}$2(2+5)2
Area=$7$7 $units^2$units2
Consider the function drawn below:
Calculate geometrically, the area bounded by the curve and the $x$x-axis over $0\le x\le4$0≤x≤4.
Find the exact value of $\int_0^{12}f\left(x\right)dx$∫120f(x)dx geometrically, where $y=f\left(x\right)$y=f(x) is graphed below.
Find the exact value of $\int_{-6}^6\sqrt{36-x^2}dx$∫6−6√36−x2dx geometrically.
The function $f\left(x\right)$f(x) is defined as:
$2x$2x | if $0\le x\le3$0≤x≤3 | |||
$f\left(x\right)$f(x) | $=$= | $6$6 | if $3 |
|
$18-2x$18−2x | if $6\le x\le9$6≤x≤9 |
Graph $f\left(x\right)$f(x) on the axis below.
Hence, calculate geometrically the area bounded by the curve and the $x$x-axis.