Problem 41 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the definite integral. \(\int_{0}^{1} x^{2}\left(1+2 x^{3}\right)^{5} d x\)

Step-by-Step Solution

Answer

The value of the integral is \( \frac{182}{9} \).

Step 1: Identify the Method

To evaluate the given integral \( \int_{0}^{1} x^{2}(1+2x^{3})^{5} \, dx \), we notice that a substitution method is suitable because the integrand involves a composition of functions.

Step 2: Make a Substitution

Let \( u = 1 + 2x^3 \). Then, differentiate both sides to find \( du \): \( du = 6x^2 \, dx \). This implies that \( x^2 \, dx = \frac{1}{6} \, du \). Additionally, adjust the limits of integration: when \( x = 0 \), \( u = 1 \), and when \( x = 1 \), \( u = 3 \).

Step 3: Rewrite the Integral

Substitute \( u \) into the integral: \[ \int_{1}^{3} (1+2x^3)^5 x^2 \, dx = \int_{1}^{3} u^5 \cdot \frac{1}{6} \, du = \frac{1}{6} \int_{1}^{3} u^5 \, du. \]

Step 4: Evaluate the Integral

To integrate \( \int u^5 \, du \), use the power rule: \[ \int u^n \, du = \frac{u^{n+1}}{n+1} + C. \]Applying this, we get: \[ \int u^5 \, du = \frac{u^6}{6}. \]

Step 5: Compute the Definitive Integral

Now, substitute and simplify: \[ \frac{1}{6} \left[ \frac{u^6}{6} \right]_{1}^{3} = \frac{1}{36} \left(3^6 - 1^6\right). \] Calculate \( 3^6 = 729 \) and \( 1^6 = 1 \). So, the integral becomes: \[ \frac{1}{36} (729 - 1) = \frac{728}{36} = \frac{182}{9}. \]

Step 6: Conclusion

The evaluated integral is \( \frac{182}{9} \).

Key Concepts

Substitution Method

The substitution method in calculus is a powerful technique used to simplify integration, especially when dealing with complex functions. The idea is to transform the original integral into a new one that is easier to evaluate by using a substitution. For example, consider the integral \( \int x^2(1+2x^3)^5 \, dx \).

• Identify a simpler variable, usually denoted as \( u \), that represents part of the integrand.
• Differentiating this new variable \( u \) helps us find \( du \), which is related to \( dx \).

For this integral, we set \( u = 1 + 2x^3 \). Differentiating gives \( du = 6x^2 \, dx \), simplifying \( x^2 \, dx \) to \( \frac{1}{6} \, du \). This transformation allows us to reframe the original problem into a more manageable form.

Integrals in Calculus

Integrals in calculus represent the concept of accumulation, such as finding the area under a curve or solving some differential equations. There are two main types of integrals: indefinite and definite. A definite integral not only deals with a function but also includes upper and lower limits, offering a specific numerical value.
When evaluating the definite integral \( \int_{0}^{1} x^2(1+2x^3)^5 \, dx \), we aim to find the area bounded by the graph of the function from \( x = 0 \) to \( x = 1 \). The process involves rewriting the problem using the substitution method and calculating within the new limits introduced by the substitution. This shifts the challenge from computing a complex integral to dealing with simpler algebraic expressions.

Power Rule for Integration

The power rule for integration is an essential tool in calculus, allowing us to integrate functions of the form \( x^n \). This rule states that for any real number \( n eq -1 \),
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \] where \( C \) is the constant of integration for indefinite integrals.
For definite integrals, we apply the power rule and then use the limits of integration to compute the final value. In our problem, after substituting \( u = 1 + 2x^3 \) and simplifying, we have \( \int u^5 \, du \). The power rule helps simplify this to \( \frac{u^6}{6} \), which is then evaluated over the limits \( u =1 \) to \( u =3 \). This yields the integral's numerical value, typically without the constant of integration in definite integrals.

Changing Limits of Integration

One of the unique steps in using the substitution method for definite integrals is changing the limits of integration.
When a substitution is made, such as \( u = 1 + 2x^3 \), the original limits \( x = 0 \) and \( x = 1 \) must be converted to the \( u \)-values using the substitution equation.

• Evaluate \( u \) at the original lower limit: when \( x = 0 \), \( u = 1 + 2(0)^3 = 1 \).
• Evaluate \( u \) at the original upper limit: when \( x = 1 \), \( u = 1 + 2(1)^3 = 3 \).

This changes the integration limits from \( 0 \) and \( 1 \) to \( 1 \) and \( 3 \) for the new integral in terms of \( u \). Using these adjusted limits is crucial for correctly evaluating the integral and obtaining the final result of \( \frac{182}{9} \). This step ensures the integrity of the calculation when transitioning between variables.