Question Statement

Find the area of the region bounded above by $y=x+6$, bounded below by $y=x^2$ and bounded on the sides by the lines $x=0$ and $x=2$.

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Background and Explanation

This problem requires calculating the area between two curves using definite integration. When a region is bounded above by $f(x)$ and below by $g(x)$ over an interval $[a,b]$, the area is given by integrating the difference $f(x)-g(x)$ between the specified limits.

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Solution

First, identify the bounding functions:

• Upper function: $f(x)=x+6$
• Lower function: $g(x)=x^2$
• Vertical bounds: $x=0$ to $x=2$

The area between curves is given by the formula: $\text{Area}={\int }_0^2(f(x)-g(x))dx$

Substituting the functions and evaluating step-by-step:

$\begin{array}{rl}& ={\int }_0^2\left(x+6-x^2\right)dx\\ & ={\int }_0^{2}xdx+6{\int }_0^{2}1dx-{\int }_0^2{x}^{2}dx\\ & ={\left|\frac{x^2}{2}\right|}_0^2+6|x{|}_0^2-{\left|\frac{x^3}{3}\right|}_0^2\\ & =\frac{1}{2}\left((2{)}^2-0\right)+6(2-0)-\frac{1}{3}\left((2{)}^3-(0{)}^3\right)\\ & =\frac{1}{2}(4-0)+6(2)-\frac{1}{3}(8-0)\\ & =2+12-\frac{8}{3}\\ & =14-\frac{8}{3}\\ & =\frac{42-8}{3}=\frac{34}{3}\text{ sq. unit}\end{array}$

Here, the vertical bar notation ${\left|F(x)\right|}_a^b$ denotes the evaluation $F(b)-F(a)$ according to the Fundamental Theorem of Calculus.

Therefore, the area of the region is $\frac{34}{3}$ square units.

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Key Formulas or Methods Used

• Area between curves: $A={\int }_a^b[f(x)-g(x)]dx$ where $f(x)$ is the upper function and $g(x)$ is the lower function over $[a,b]$
• Power rule for integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for $n\ne -1$
• Linearity of integration: $\int [\alpha f(x)+\beta g(x)]dx=\alpha \int f(x)dx+\beta \int g(x)dx$
• Fundamental Theorem of Calculus: ${\int }_a^{b}f(x)dx=F(b)-F(a)$ where $F$ is an antiderivative of $f$

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Summary of Steps

1. Identify the upper function ($y=x+6$), lower function ($y=x^2$), and interval $[0,2]$
2. Set up the definite integral: ${\int }_0^2[(x+6)-x^2]dx$
3. Separate into individual integrals using the linearity property
4. Find the antiderivatives: $\frac{x^2}{2}$, $6x$, and $\frac{x^3}{3}$
5. Evaluate each antiderivative at the upper bound ($x=2$) and lower bound ($x=0$), then subtract
6. Simplify the resulting arithmetic expression: $2+12-\frac{8}{3}=\frac{34}{3}$
7. State the final answer with appropriate units: $\frac{34}{3}$ square units