Problem 63 - Calculus Early Transcendentals

Question

Evaluate the following definite integrals. Use Theorem 10 to express your answer in terms of logarithms. \(\int_{1 / 8}^{1} \frac{d x}{x \sqrt{1+x^{2 / 3}}}\)

Step-by-Step Solution

Answer

Question: Evaluate the definite integral: \(\int_{1 / 8}^{1} \frac{1}{x \sqrt{1+x^{2 / 3}}} dx\). Answer: 3(log(1+\sqrt{2}) - log(\frac{1}{4} + \sqrt{\frac{1}{16} + 1}))

Step 1: Identify the function to integrate and the interval

The function to integrate is \(f(x) = \frac{1}{x\sqrt{1+x^{2/3}}}\) and the interval is from \(a=\frac{1}{8}\) to \(b=1\).

Step 2: Apply 'u' substitution

We can do a 'u' substitution for this integral: let \(u = x^{2/3}\), then \(x = u^{3/2}\), and \(dx = \frac{3}{2}u^{1/2}du\). Now plug in the expressions into the original integral: \(\int_{1 / 8}^{1} \frac{d x}{x \sqrt{1+x^{2 / 3}}}= \int_{1}^{64} \frac{\frac{3}{2}u^{1/2} du}{u^{3/2} \sqrt{1+u}}\).

Step 3: Simplify the integral

Now, simplify the integral: \(\int_{1}^{64} \frac{\frac{3}{2}u^{1/2} du}{u^{3/2} \sqrt{1+u}} = \frac{3}{2} \int_{1}^{64} \frac{1}{\sqrt{u} \sqrt{1+u}} du\)

Step 4: Identify the antiderivative

Integrating the simplified expression, we get the antiderivative: \(F(u) = \frac{3}{2} \int \frac{1}{\sqrt{u} \sqrt{1+u}} du = 3 sinh^{-1}(\sqrt{u}) + C\). (Note: sinh^{-1}(z) is the inverse hyperbolic sine function)

Step 5: Reverse 'u' substitution

We now need to reverse the substitution from step 2. Replace u with \(x^{2/3}\): \(F(x) = 3 sinh^{-1}(\sqrt{x^{2/3}}) + C\)

Step 6: Apply Theorem 10

Now, apply the Fundamental Theorem of Calculus using the antiderivative and the given interval: \(\int_{1 / 8}^{1} \frac{1}{x \sqrt{1+x^{2 / 3}}} dx = F(1)-F(\frac{1}{8}) = 3 sinh^{-1}(\sqrt{1^{2/3}}) - 3 sinh^{-1}(\sqrt{(\frac{1}{8})^{2/3}}) = 3 sinh^{-1}(1) - 3 sinh^{-1}(\frac{1}{4})\)

Step 7: Simplify the result

Finally, simplify the result: \(3 sinh^{-1}(1) - 3 sinh^{-1}(\frac{1}{4}) = 3(log(1 + \sqrt{2}) - log(\frac{1}{4} + \sqrt{\frac{1}{4}^2 + 1}))\) The final answer is: \(= 3(log(1+\sqrt{2}) - log(\frac{1}{4} + \sqrt{\frac{1}{16} + 1})\)

Key Concepts

u-substitution

U-substitution is a helpful technique to simplify integrals, especially when dealing with complex functions. The process involves changing variables to make the integral easier to evaluate.

In this exercise, we use this technique by letting \( u = x^{2/3} \). This substitution transforms the variable, making the integration process more manageable. As a result, \( x \) can be expressed as \( u^{3/2} \) and the differential \( dx \) can be rewritten as \( \frac{3}{2}u^{1/2}du \).

This substitution changes the original integral into one with a new variable, \( u \), and adjusts the limits of integration accordingly, from \( x = 1/8 \) to \( x = 1 \) and thus, \( u = 1 \) to \( u = 64 \).

The essence of u-substitution is to transform complex integrals into a simpler form, allowing one to find the antiderivative with ease.

inverse hyperbolic functions

Inverse hyperbolic functions play a vital role in calculus, especially when integrating certain functions. They are the inverses of the hyperbolic functions, like \( \sinh(x) \), \( \cosh(x) \), and \( \tanh(x) \).

In solving this integral, we arrived at an expression involving the inverse hyperbolic sine function, \( \sinh^{-1}(x) \). This function helps express antiderivatives that would otherwise be difficult to represent using standard functions.

The antiderivative, \( 3 \sinh^{-1}(\sqrt{u}) + C \), provides a solution with the help of inverse hyperbolic functions, which relate to logarithms. This approach is efficient and crucial, especially for typical calculus problems involving square roots and other non-linear transformations.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration and is essential for evaluating definite integrals. It states that if \( F \) is an antiderivative of \( f \) over an interval \([a,b]\), then:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

This theorem effectively allows us to calculate the value of a definite integral using an antiderivative.

In the solution, we used this theorem after computing the antiderivative \( 3 \sinh^{-1}(\sqrt{x^{2/3}}) + C \). By substituting the boundary values \( x = 1 \) and \( x = 1/8 \), we found the specific value for the integral:

• \( F(1) - F(1/8) = 3 \sinh^{-1}(1) - 3 \sinh^{-1}(\frac{1}{4}) \)

Applying the theorem simplifies calculations and provides a direct way to determine the area under a curve, making it invaluable for calculus students.