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Chapter 5: Problem 53

51-70 Evaluate the definite integral. \(\int_{0}^{1} x^{2}\left(1+2 x^{3}\right)^{5} d x\)

Short Answer

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

Step by step solution

01

Choose Substitution

We start by choosing an appropriate substitution. Let's set \( u = 1 + 2x^3 \). This implies that we will be rewriting the integral in terms of \( u \).
02

Differentiate the Substitution

Differentiate \( u = 1 + 2x^3 \) with respect to \( x \) to find \( du \). We get \( \frac{du}{dx} = 6x^2 \), which implies \( du = 6x^2 \, dx \).
03

Express \(dx\) in Terms of \(du\)

We solve for \( x^2 \, dx \) from \( du = 6x^2 \, dx \). This gives us \( x^2 \, dx = \frac{1}{6} du \).
04

Change the Limits of Integration

When \( x = 0 \), \( u = 1 + 2(0)^3 = 1 \). When \( x = 1 \), \( u = 1 + 2(1)^3 = 3 \). So the limits change from \([0, 1]\) to \([1, 3]\).
05

Substitute and Simplify the Integral

Substitute \( u \) and \( x^2 \, dx \) into the integral to get \( \int_{1}^{3} \frac{1}{6} u^{5} \, du \). This simplifies to \( \frac{1}{6} \int_{1}^{3} u^{5} \, du \).
06

Compute the Antiderivative

The antiderivative of \( u^5 \) is \( \frac{u^6}{6} \). Therefore, \( \int u^5 \, du = \frac{u^6}{6} + C \).
07

Evaluate the Definite Integral

Evaluate \( \frac{1}{6} \left[ \frac{u^6}{6} \right] \) from 1 to 3. This becomes \( \frac{1}{36} [3^6 - 1^6] \).
08

Simplify the Final Expression

Calculate \( 3^6 = 729 \) and \( 1^6 = 1 \). Then \( 729 - 1 = 728 \). Evaluate \( \frac{1}{36} \times 728 = \frac{728}{36} \). Simplify by dividing both numerator and denominator by 4, which gives \( \frac{182}{9} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
When solving a definite integral problem, choosing the right substitution can streamline the process significantly. In our example, we set \( u = 1 + 2x^3 \). This approach converts the integral into a form that is easier to work with.
Here's why substitution is essential:
  • It simplifies the function you are integrating, making calculations more manageable.
  • It transforms complex algebra into simpler terms.
In this particular example, by choosing \( u = 1 + 2x^3 \), it secures a new variable, \( u \), that simplifies the expression within the integrand. We also calculate \( \frac{du}{dx} = 6x^2 \), leading us to \( du = 6x^2 \, dx \). This relationship allows us to express \( x^2 \, dx \) in terms of \( du \), which is essential for simplifying the integral.
Antiderivative Calculation
Calculating the antiderivative is a critical step in solving definite integrals. It transforms the function back to its original form.
Our new integral, after substitution, is \( \int_{1}^{3} \frac{1}{6} u^{5} \, du \).
Here's the process to find the antiderivative:
  • We identify the antiderivative of \( u^5 \) as \( \frac{u^6}{6} \).
  • The constant \( \frac{1}{6} \) in front is carried along throughout the calculation.
  • This results in \( \frac{1}{6} \cdot \frac{u^6}{6} = \frac{u^6}{36} \).
The antiderivative is now in a form that allows us to evaluate it between the limits of integration, ensuring we can find the area under the curve defined by the original integral.
Integration Limits
Integration limits define the portion of the curve you are considering. Changing them correctly after substitution is crucial.
For our problem, the original limits are \( x = 0 \) and \( x = 1 \). Here's how they change:
  • For \( x = 0 \), substitute into \( u = 1 + 2x^3 \) to get \( u = 1 \).
  • For \( x = 1 \), substitute similarly to find \( u = 3 \).
Now the new limits are from \( u = 1 \) to \( u = 3 \), which align with our new integral: \( \int_{1}^{3} \frac{1}{6} u^{5} \, du \).
Evaluating this integral within these adjusted limits helps calculate the exact definite integral, giving us the total accumulated value between these points.

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