Question Statement

Evaluate the definite integral:

${\int }_{-4}^{12}\sqrt{y+4}dy$

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Background and Explanation

This problem involves integrating a radical function by first converting it to rational exponent form, then applying the power rule for integration. The key is recognizing that $\sqrt{y+4}=(y+4{)}^{1/2}$ and that the derivative of the inner function $(y+4)$ is simply 1, allowing direct application of the power rule without additional substitution.

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Solution

Begin by rewriting the square root as a power with a rational exponent to prepare for integration:

$\sqrt{y+4}=(y+4{)}^{1/2}$

Apply the power rule for integration $\int u^{n}du=\frac{u^{n+1}}{n+1}$ where $n=\frac{1}{2}$:

$\begin{array}{rl}{\int }_{-4}^{12}\sqrt{y+4}dy& ={\int }_{-4}^{12}(y+4{)}^{1/2}\cdot 1dy\\ & ={\left|\frac{(y+4{)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right|}_{-4}^{12}\\ & ={\left|\frac{(y+4{)}^{\frac{3}{2}}}{\frac{3}{2}}\right|}_{-4}^{12}\end{array}$

Simplify the complex fraction by multiplying by the reciprocal of the denominator:

$\begin{array}{rl}& =\frac{2}{3}{\left|(y+4{)}^{3/2}\right|}_{-4}^{12}\end{array}$

Evaluate the antiderivative at the upper limit ($y=12$) and lower limit ($y=-4$) using the Fundamental Theorem of Calculus:

$\begin{array}{rl}& =\frac{2}{3}\left[(12+4{)}^{3/2}-(-4+4{)}^{3/2}\right]\\ & =\frac{2}{3}\left[(16{)}^{3/2}-(0{)}^{3/2}\right]\end{array}$

Simplify $(16{)}^{3/2}$ by recognizing that $16=4^2$ and applying exponent rules $(a^m{)}^n=a^{mn}$:

$(16{)}^{3/2}={\left(4^2\right)}^{3/2}=4^3=64$

Substitute this value back and complete the calculation:

$\begin{array}{rl}& =\frac{2}{3}\left[64-0\right]\\ & =\frac{2}{3}(64)\\ & =\frac{128}{3}\end{array}$

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Key Formulas or Methods Used

• Power Rule for Integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Rational Exponents: $\sqrt[n]{x}=x^{1/n}$
• Exponent Rules: $(a^m{)}^n=a^{mn}$
• Fundamental Theorem of Calculus: ${\int }_a^{b}f(y)dy=F(b)-F(a)$ where $F$ is the antiderivative of $f$

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Summary of Steps

1. Rewrite the integrand using rational exponents: $\sqrt{y+4}=(y+4{)}^{1/2}$
2. Apply the power rule to find the antiderivative: $\frac{(y+4{)}^{3/2}}{3/2}=\frac{2}{3}(y+4{)}^{3/2}$
3. Evaluate the antiderivative at the upper bound ($y=12$) and lower bound ($y=-4$)
4. Calculate the powers: $(16{)}^{3/2}=64$ and $(0{)}^{3/2}=0$
5. Compute the final result: $\frac{2}{3}(64-0)=\frac{128}{3}$