This problem aims to evaluate the integrals given against the graph $g$. The concept behind this problem is related to definite integration and calculating the area under the curve, which is basically another definition of integration.
The area under a curve of two points is computed by taking a definite integral between those two points.
Let’s say you want to find the area under the curve $y = f(x)$ which lies between $x = a$ and $x = b$, you have to integrate $y = f(x)$ between the given limits of $a$ and $b$.
Expert Answer
We are given $3$ different integrals, each representing a shape or a line in the given graph. We will start by evaluating each integral one by one.
Part a:
\[\int^{6}_{0} g(x)\space dx\]
If we look at the graph we see that on the interval $[0, 2]$, the graph is just a straight line that comes down from $y = 12$ to $y = 0$. If you look closely this straight line represents a triangle along the $y$ axis as its perpendicular.
Thus the area of this portion is just the area of the triangle, whose base is $6$ and has a height of $12$ units. So calculating the area:
\[=\dfrac{1}{2}\cdot b\cdot h\]
\[=\dfrac{1}{2}\cdot 6\cdot 12\]
\[=36\]
Since the area lies above the $x$ axis, so $\int^{6}_{0} g(x)\space dx$ equals the area.
Hence, $\int^{6}_{0} g(x)\space dx=36$.
Part b:
\[\int^{18}_{0} g(x)\space dx\]
On the interval $[6, 18]$, the graph is just a semi-circle below the $x$ axis that has a radius of $6$ units.
Thus it’s a semi-circle, with a radius of $6$ units. So calculating the area:
\[=\dfrac{1}{2}\cdot \pi\cdot r^2\]
\[=\dfrac{1}{2}\cdot \pi\cdot 6^2\]
\[=\dfrac{1}{2}\cdot \pi\cdot 36\]
\[=18\pi\]
Since the area lies below the $x$ axis, so the integral would have a negative sign. And $\int^{18}_{6} g(x)\space dx$ equals the area.
Hence, $\int^{18}_{6} g(x)\space dx=-18\pi$.
Part c:
\[\int^{21}_{0} g(x)\space dx\]
We can rewrite the above integral as:
\[\int^{21}_{0} g(x)\space dx = \int^{6}_{0} g(x)\space dx + \int^{18}_{6} g(x)\space dx + \int^{21}_{18} g(x)\space dx\]
This gives us:
\[=36 – 18\pi + \int^{21}_{18} g(x)\space dx\]
So we just have to calculate the integral $\int^{21}_{18} g(x)\space dx$.
On the interval $[18, 21]$, the graph is a straight line that goes up from $y = 0$ to $y = 3$. This straight line represents a triangle with a base of $3$ and a height of $3$ units. So calculating the area:
\[=\dfrac{1}{2}\cdot 3\cdot 3\]
\[=\dfrac{9}{2}\]
Since the area lies above the $x$ axis, so $\int^{21}_{18} g(x)\space dx=\dfrac{9}{2}$.
Hence,
\[\int^{21}_{0} g(x)\space dx=36-18\pi+\dfrac{9}{2}=-16.05\]
Numerical Results
Part a: $\int^{6}_{0} g(x)\space dx=36$
Part b: $\int^{18}_{6} g(x)\space dx=-18\pi$
Part c: $\int^{21}_{0} g(x)\space dx=-16.05$
Example
For the given function $f(x) = 7 – x^2$, calculate the area under the curve with limits $x = -1$ to $2$.
The area under the curve can be calculated as:
\[ = \int^{2}_{-1} f(x)\space dx \]
\[ = \int^{2}_{-1} (7 – x^2)\space dx \]
\[= (7x – \dfrac{1}{3}x^3)|^{2}_{-1}\]
\[= [7\cdot 2 – \dfrac{1}{3}(8)]- [7(-1) – \dfrac{1}{3}(-1)] \]
\[= [\dfrac{(42-8)}{3}]- [\dfrac{1-21}{3}]\]
\[= \dfrac{(54)}{3}\]
\[= 18 sq.units \]