banner

The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral.

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:

\[ = \int^{2}_{-1} f(x)\space dx \]

\[ = \int^{2}_{-1} (7 – x^2)\space dx \]

\[= (7x – \dfrac{1}{3}x^3)|^{2}_{-1}\]

\[= [7\cdot 2 – \dfrac{1}{3}(8)]- [7(-1) – \dfrac{1}{3}(-1)] \]

\[= [\dfrac{(42-8)}{3}]- [\dfrac{1-21}{3}]\]

\[= \dfrac{(54)}{3}\]

\[= 18 sq.units \]

5/5 - (7 votes)