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:

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

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

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

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

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

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

\[= 18 sq.units \]