Question Statement

Evaluate the definite integral:

${\int }_0^{\frac{1}{2}}(2x+1{)}^{-\frac{1}{3}}dx$

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Background and Explanation

This problem requires integrating a function of the form $(ax+b{)}^n$, which follows the generalized power rule for integration. When the base is a linear expression $(2x+1)$, we must account for the coefficient of $x$ (which is 2) when applying the antiderivative formula.

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Solution

To evaluate this integral, we use the power rule for integration adapted for linear expressions inside the power function.

Step 1: Prepare the integral for substitution

We need the derivative of the inner function $(2x+1)$, which is $2$, to appear in the numerator. We multiply and divide by 2 to maintain equality:

$\begin{array}{rl}{\int }_0^{\frac{1}{2}}(2x+1{)}^{-\frac{1}{3}}dx& =\frac{1}{2}{\int }_0^{1/2}(2x+1{)}^{-1/3}\cdot 2dx\end{array}$

Step 2: Apply the power rule

Using the formula $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C$ with $a=2$ and $n=-\frac{1}{3}$:

$\begin{array}{rl}& =\frac{1}{2}{\left[\frac{(2x+1{)}^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\right]}_0^{1/2}\\ & =\frac{1}{2}{\left[\frac{(2x+1{)}^{2/3}}{\frac{2}{3}}\right]}_0^{1/2}\end{array}$

Step 3: Simplify the constants

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

$\begin{array}{rl}& =\frac{1}{2}\cdot \frac{3}{2}{\left[(2x+1{)}^{2/3}\right]}_0^{1/2}\\ & =\frac{3}{4}{\left[(2x+1{)}^{2/3}\right]}_0^{1/2}\end{array}$

Step 4: Evaluate at the limits

Substitute the upper limit $x=\frac{1}{2}$ and lower limit $x=0$:

$\begin{array}{rl}& =\frac{3}{4}\left[{\left(2\left(\frac{1}{2}\right)+1\right)}^{2/3}-(2(0)+1{)}^{2/3}\right]\\ & =\frac{3}{4}\left[(1+1{)}^{2/3}-(0+1{)}^{2/3}\right]\\ & =\frac{3}{4}\left[2^{2/3}-1\right]\end{array}$

Therefore, the value of the integral is $\boxed{\frac{3}{4}\left(2^{2/3}-1\right)}$.

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Key Formulas or Methods Used

• Generalized Power Rule for Integration: $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C$ (for $n\ne -1$)
• Definite Integral Evaluation: ${\int }_a^{b}f(x)dx=F(b)-F(a)$ where $F$ is the antiderivative of $f$
• Adjustment for Chain Rule: When the inner function has coefficient $a$, multiply by $\frac{1}{a}$ outside the integral to compensate for the factor $a$ needed in the numerator

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Summary of Steps

1. Adjust for the chain rule: Multiply and divide by the coefficient of $x$ (which is 2) to prepare for integration
2. Increase the power by 1: Apply the power rule by adding 1 to the exponent $-\frac{1}{3}$ to get $\frac{2}{3}$
3. Divide by the new power: Divide by $\frac{2}{3}$ (or multiply by $\frac{3}{2}$) and account for the coefficient 2 from the chain rule adjustment
4. Simplify constants: Combine $\frac{1}{2}\times \frac{3}{2}=\frac{3}{4}$ outside the brackets
5. Evaluate at bounds: Substitute $x=\frac{1}{2}$ and $x=0$, then subtract to get the final numerical result $\frac{3}{4}(2^{2/3}-1)$