Problem 13 - Calculus: One and Several Variables

Question

Calculate. $$\int_{0}^{5} x^{2} \sqrt{25-x^{2}} d x$$.

Step-by-Step Solution

Answer

Thus, the definite integral \(\int_{0}^{5} x^{2} \sqrt{25-x^{2}} dx\) evaluates to \(\frac{125}{3}\).

Step 1: Identify the substitution

Observe that the function inside the square root is of the form \(25-x^2\). So, let's make a substitution: Let \(u = 25 - x^2\), then differentiate with respect to x: \(\frac{du}{dx} = -2x\). Now solve for dx: \(dx = \frac{du}{-2x}\).

Step 2: Rewrite the integral using the substitution

Now that we have our substitution and our expression for dx, let's rewrite the integral: \(\int_{0}^{5} x^{2} \sqrt{25-x^{2}} dx = \int x^2 \sqrt{u} \frac{du}{-2x}\). Notice that the \(x\) in the numerator cancels out with the \(x\) in the denominator, so the integral simplifies to: \(-\frac{1}{2}\int \sqrt{u} du\).

Step 3: Change the limits of integration

Since we made a substitution, we need to change the limits of integration to match the variable u. We have: For x = 0: \(u = 25 - 0^2 = 25\), For x = 5: \(u = 25 - 5^2 = 0\). So, our integral is now \(-\frac{1}{2}\int_{25}^{0} \sqrt{u} du\).

Step 4: Evaluate the integral

To evaluate the integral, we use the power rule for integration: \(\int u^n du = \frac{u^{n+1}}{n+1} + C\). Our integral becomes: \(-\frac{1}{2}\int_{25}^{0} u^{\frac{1}{2}} du = -\frac{1}{2}\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_{25}^{0}\). Now we need to plug in the limits of integration and subtract: \(-\frac{1}{2}\left(\frac{0^{\frac{3}{2}}}{\frac{3}{2}} - \frac{25^{\frac{3}{2}}}{\frac{3}{2}}\right) = \frac{1}{3}(25\sqrt{25})\).

Step 5: Simplify the answer

Our final answer is: \(\frac{1}{3}(25\sqrt{25}) = \frac{1}{3}(25\cdot 5) = \frac{125}{3}\). Therefore, \(\int_{0}^{5} x^{2} \sqrt{25-x^{2}} dx = \frac{125}{3}\).

Key Concepts

Definite Integral

A definite integral represents the accumulated area under a curve between two points on the x-axis. It is written in the form \( \int_{a}^{b} f(x) dx \) where \(a\) and \(b\) are the limits of integration (also known as the bounds of integration), and \(f(x)\) is the function to integrate. In the context of the given exercise, the definite integral \( \int_{0}^{5} x^{2} \sqrt{25-x^{2}} dx \) is looking for the area under the curve of \(x^{2}\sqrt{25-x^{2}}\) from x = 0 to x = 5.

The solution process involves evaluating this area by finding the antiderivative of the function and then applying the limits of integration which will reveal the specific value of the area encompassed by these bounds.

Power Rule for Integration

The power rule for integration is an essential concept in calculus and serves as a basic technique for integrating powers of \(x\). The rule states that for any real number \(neq-1\), the integral of \(x^n\) with respect to \(x\) is given by \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration. In our case, with a square root function, we can think of \(\sqrt{u}\) as \(u^{1/2}\), applying the power rule results in \( \frac{u^{(1/2 + 1)}}{(1/2 + 1)} \) or simplified as \( \frac{u^{3/2}}{3/2} \) when integrating. This application simplifies the process of integrating the square root function after performing the u-substitution method.

Change of Limits in Integral

When using substitution in definite integrals, it's important to change the limits of integration to align with the new variable. Otherwise, you would need to revert back to the original variable after integrating. For instance, if you substitute \(u = 25 - x^2\), the limits for \(x\) initially at 0 and 5 must be converted to new limits for \(u\). As shown in the given exercise, when \(x = 0\), \(u = 25 - (0)^2 = 25\), and when \(x = 5\), \(u = 25 - (5)^2 = 0\). This leads to the new transformed integral running from 25 to 0. Notably, if the limits were not changed, one would have to substitute back to \(x\) after integrating in terms of \(u\) before applying the fundamental theorem of calculus to find the definite integral.

U-Substitution Method

U-substitution is a technique used in integration to simplify certain types of integrals. It is analogous to the chain rule for differentiation. Essentially, it involves substituting a part of the integrand that allows for easier integration. This is accomplished by choosing a new variable, \(u\), which is a function of \(x\), then expressing \(dx\) in terms of \(du\).

In this solution, setting \(u = 25 - x^2\) and expressing \(dx\) as \(du/{-2x}\) allows us to cancel out an \(x\) and integrate a simpler function of \(u\). After integrating with respect to \(u\), we often simplify the expression and then back-substitute to express our answer in terms of the original variable, \(x\), unless we have changed the limits of integration to \(u\), as we did in the exercise presented.