Question Statement

Find the area under the curve $y=\sqrt{6x+4}$ (above the $x$-axis) from $x=0$ to $x=2$.

---

Background and Explanation

To find the area under a curve $y=f(x)$ between two points, we evaluate the definite integral ${\int }_a^{b}f(x)dx$. This problem requires integrating a radical function using the power rule and the reverse chain rule (or substitution method).

---

Solution

The area under a curve is calculated by integrating the function over the given interval.

Setting up the Integral

$\text{Area}={\int }_0^{2}ydx={\int }_0^2\sqrt{6x+4}dx$

Evaluating the Integral

We rewrite the integrand to prepare for integration using the power rule. Notice that the derivative of the inner function $(6x+4)$ is $6$, so we multiply and divide by $6$:

$=\frac{1}{6}{\int }_0^2(6x+4{)}^{1/2}\cdot 6dx$

Now apply the power rule for integration $\int u^{n}du=\frac{u^{n+1}}{n+1}$ where $u=6x+4$ and $n=\frac{1}{2}$:

$=\frac{1}{6}\cdot {\left[\frac{(6x+4{)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^2$

Simplify the exponent and denominator:

$=\frac{1}{6}\cdot {\left[\frac{(6x+4{)}^{3/2}}{3/2}\right]}_0^2$

Dividing by $\frac{3}{2}$ is equivalent to multiplying by $\frac{2}{3}$:

$=\frac{1}{6}\cdot \frac{2}{3}{\left[(6x+4{)}^{3/2}\right]}_0^2$

Simplify the constant coefficient $\frac{1}{6}\cdot \frac{2}{3}=\frac{2}{18}=\frac{1}{9}$:

$=\frac{1}{9}\left[(6(2)+4{)}^{3/2}-(6(0)+4{)}^{3/2}\right]$

Simplifying the Expression

Evaluate the expressions inside the brackets:

$=\frac{1}{9}\left[(12+4{)}^{3/2}-(0+4{)}^{3/2}\right]$

$=\frac{1}{9}\left[(16{)}^{3/2}-(4{)}^{3/2}\right]$

Simplify the fractional exponents using the property $(a^m{)}^n=a^{mn}$:

• $(16{)}^{3/2}=(4^2{)}^{3/2}=4^3=64$
• $(4{)}^{3/2}=(2^2{)}^{3/2}=2^3=8$

Therefore:

$=\frac{1}{9}\left[64-8\right]$

$=\frac{1}{9}\cdot 56$

$=\frac{56}{9}\text{ square units}$

---

Key Formulas or Methods Used

• Area under curve: $\text{Area}={\int }_a^{b}f(x)dx$
• Power rule for integration: $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C$ for $n\ne -1$
• Reverse chain rule: Adjusting for the derivative of the inner linear function by multiplying/dividing by the coefficient of $x$
• Exponent rules: $(a^m{)}^n=a^{mn}$ and $a^{m/n}=\sqrt[n]{a^m}$

---

Summary of Steps

1. Set up the definite integral: ${\int }_0^2\sqrt{6x+4}dx$
2. Prepare for integration by writing as $\frac{1}{6}{\int }_0^2(6x+4{)}^{1/2}\cdot 6dx$
3. Apply power rule: $\frac{1}{6}\cdot \frac{2}{3}(6x+4{)}^{3/2}$ evaluated from $0$ to $2$
4. Simplify coefficient to $\frac{1}{9}$ and substitute limits: $\frac{1}{9}[(16{)}^{3/2}-(4{)}^{3/2}]$
5. Evaluate powers: $16^{3/2}=64$ and $4^{3/2}=8$
6. Calculate final result: $\frac{1}{9}(56)=\frac{56}{9}$ square units