Question

Evaluate \(\int_{0}^{1}\left(\sqrt[3]{1-x^{7}}-\sqrt[7]{1-x^{3}}\right) d x\)

Short answer

Question: Evaluate the integral \(\int_{0}^{1}\left(\sqrt[3]{1-x^{7}}-\sqrt[7]{1-x^{3}}\right) dx\) Answer: \(\frac{2}{3}\)

Step-by-step solution

Split the integral into two separate integrals

Split the given integral into two separate integrals: $$\int_{0}^{1}\left(\sqrt[3]{1-x^{7}}-\sqrt[7]{1-x^{3}}\right) dx = \int_{0}^{1}\sqrt[3]{1-x^{7}} dx - \int_{0}^{1}\sqrt[7]{1-x^{3}} dx$$

Find the antiderivatives for each integral

We will use the substitution method to find the antiderivatives of each part For the first integral: Let \(u = 1-x^7\) then \(du = -7x^6 dx\) and \(dx = \frac{1}{-7x^6} du\) $$\int_{0}^{1}\sqrt[3]{1-x^{7}} dx = \int_{1}^{0}\sqrt[3]{u} \frac{1}{-7x^6} du$$ For the second integral: Let \(v = 1-x^3\) then \(dv = -3x^2 dx\) and \(dx = \frac{1}{-3x^2} dv\) $$\int_{0}^{1}\sqrt[7]{1-x^{3}} dx = \int_{1}^{0}\sqrt[7]{v} \frac{1}{-3x^2} dv$$

Evaluate the integrals with the new substitutions

Evaluate each integral with the new variable substitutions: First integral: $$\int_{1}^{0}\sqrt[3]{u} \frac{1}{-7(-u)^{\frac{6}{7}}} du= \frac{-1}{7}\int_{1}^{0}u^{\frac{1}{3}}(-1)u^{-\frac{6}{7}}du = \frac{1}{7}\int_{1}^{0}u^{(\frac{1}{3}-\frac{6}{7})}du$$ Second integral: $$\int_{1}^{0}\sqrt[7]{v} \frac{1}{-3(-v)^{\frac{2}{3}}} dv= \frac{-1}{3}\int_{1}^{0}v^{\frac{1}{7}}(-1)v^{-\frac{2}{3}}dv = \frac{1}{3}\int_{1}^{0}v^{(\frac{1}{7}-\frac{2}{3})} dv$$

Find the antiderivatives and evaluate them at the endpoints of the interval

Find the antiderivatives of these new integrals and evaluate them at the given endpoints: First integral: $$\frac{1}{7}\int_{1}^{0}u^{-\frac{14}{21}}du = \frac{1}{7}\left[\frac{-21}{\frac{-14}{21}+1}(u^{\frac{-14}{21}+1})\right]_{1}^{0} = \frac{-1}{3}\left[u^{-\frac{7}{21}}\right]_{1}^{0} = \frac{-1}{3}\left[0 - 1\right] = \frac{1}{3}$$ Second integral: $$\frac{1}{3}\int_{1}^{0}v^{-\frac{5}{21}} dv = \frac{1}{3}\left[\frac{21}{\frac{-5}{21}+1}(v^{\frac{-5}{21}+1})\right]_{1}^{0} = \frac{1}{3}\left[v^{\frac{16}{21}}\right]_{1}^{0} = \frac{1}{3}\left[0 - 1\right] = \frac{-1}{3}$$

Evaluate the original integral

Now that we have evaluated and combined the two integrals, we can find the value of the original integral: $$\int_{0}^{1}\left(\sqrt[3]{1-x^{7}}-\sqrt[7]{1-x^{3}}\right) dx = \frac{1}{3} - (-\frac{1}{3}) = \frac{2}{3}$$ Thus, the value of the given integral is \(\boxed{\frac{2}{3}}\).

Key concepts

Substitution Method

Understanding the substitution method is crucial when tackling integrals that do not fit elementary patterns. In essence, this technique simplifies integration by transforming a difficult-to-integrate function into a simpler one with a straightforward antiderivative.

This method, sometimes referred to as 'u-substitution', involves identifying a portion of the integrand that can be represented by a new variable (typically 'u' or 'v'), which allows us to rewrite the integral in terms of this new variable. The chosen part is usually a function whose derivative is present elsewhere in the integrand. After making the substitution, we replace the original differential (like 'dx') with a new differential in terms of the new variable (like 'du').

The process involves three main steps:

• Choose the substitution that will simplify the integral.
• Replace all occurrences of the original variable with the new variable and its corresponding differential.
• Integrate using the new variable, then substitute back to the original variable if necessary.

In the example exercise, we have two substitutions:

• For \(\sqrt[3]{1-x^{7}}\), we set \(u = 1-x^7\).
• For \(\sqrt[7]{1-x^{3}}\), we set \(v = 1-x^3\).

After finding the derivatives \(du\) and \(dv\) respectively, we express \(dx\) in terms of \(du\) and \(dv\) and thereby simplify the integral into a form we can manage more easily.

Antiderivatives

Antiderivatives, often called indefinite integrals, are the reverse of differentiation. For any given continuous function, finding an antiderivative means discovering a function whose derivative is the original function. It is an integral without specified limits and is essential in the process of solving definite integrals using the fundamental theorem of calculus.

When finding antiderivatives, we commonly apply rules that are the inverse of differentiation rules, such as the power rule or the exponential rule.

One key point to remember is that the antiderivative of a function is not unique – it includes a constant of integration (C), which represents an infinite family of functions.
For example, if \(F(x)\) is an antiderivative of \(f(x)\), then \(F(x) + C\), where \(C\) is any constant, is also an antiderivative of \(f(x)\).

In our exercise, after applying the substitution method, we found the antiderivatives of the resulting expressions, which were necessary for evaluating the definite integrals from step 4 onward.

Definite Integrals

Definite integrals play a central role in calculus. Unlike antiderivatives, which represent a family of functions, definite integrals are used to calculate the exact area under a curve between two points on the x-axis. They are integrals with specific upper and lower bounds, providing a number as the answer.

The process for evaluating a definite integral typically follows these steps:

• Find the antiderivative of the integrand.
• Evaluate the antiderivative at the upper limit of integration.
• Evaluate the antiderivative at the lower limit of integration.
• Subtract the value from the second step from the value from the third step.

This 'evaluate-subtract' procedure can sometimes be referred to as the Fundamental Theorem of Calculus, which connects differentiation and integration.

In our problem, after we used the substitution method to simplify the integrals, we found the antiderivatives and then evaluated them at the given limits. Finally, we subtracted the value at the lower limit from the value at the upper limit to determine the area under each curve. We then combined these to find the answer to the original definite integral problem.