Problem 23 - Calculus Single Variable

Question

Use the method of substitution to evaluate the definite integrals. $$ \int_{1}^{2}\left(t^{2}-t\right)^{5}(2 t-1) d t $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{32}{3} \).

Step 1: Choose a Substitution Variable

To use substitution, identify a part of the integration that can be replaced with a single variable. Here, we set \( u = t^2 - t \). This choice simplifies differentiation and substitution in the integral.

Step 2: Differentiate the Substitution Variable

Differentiate \( u = t^2 - t \) with respect to \( t \) to find \( du \/ dt \). Doing this, we have:\[\frac{du}{dt} = 2t - 1,\]which implies \( du = (2t - 1) \, dt \).

Step 3: Substitute in the Integral

Substitute \( u \) and \( du \) back into the integral. The original integral \( \int_1^2 (t^2 - t)^5 (2t - 1) \ dt \) becomes \( \int (u^5) \, du \) after substitution.

Step 4: Determine New Limits of Integration

Convert the original limits for \( t \) into limits for \( u \). When \( t = 1 \), \( u = 1^2 - 1 = 0 \). When \( t = 2 \), \( u = 2^2 - 2 = 2 \). Therefore, the new limits are from \( u = 0 \) to \( u = 2 \).

Step 5: Integrate with Respect to New Variable

The integral with new limits is:\[\int_0^2 u^5 \, du.\]Integrate this to get:\[\left[ \frac{u^6}{6} \right]_0^2.\]

Step 6: Evaluate the Definite Integral

Substitute the limits into the integrated expression:\[\frac{2^6}{6} - \frac{0^6}{6} = \frac{64}{6} = \frac{32}{3}.\]Thus, the value of the definite integral is \( \frac{32}{3} \).

Key Concepts

Definite Integrals

A definite integral represents the accumulation of a quantity, where the start and end points are determined by the limits of integration. It essentially calculates the net area under a curve within specific bounds, offering insights into the total accumulation between two values. When tackling definite integrals, you typically encounter two limits, such as the integral from 1 to 2 in this problem. The definite integral is evaluated by first finding an antiderivative and then computing the difference between its values at the upper and lower limits.
Consider the given problem:

• The definite integral is taken between the bounds 1 and 2, which serve as start and end points for evaluation.
• This integral is used to determine the total value of the function within these limits.

The computation involves substituting back the original variables once the definite integral has been evaluated in terms of the substitution variable, ensuring the final answer relates back to the original context.
This allows one to find meaningful outcomes within specified intervals.

Integration Techniques

Effective integration techniques simplify the process of finding the antiderivative of complex functions. The Substitution Method is a key strategy, allowing for simplification by introducing a new variable. This approach leverages the chain rule backwards by rewriting an integral in a more manageable form.
To employ the substitution method efficiently:

• Select a portion of the integrand that simplifies when substituted.
• Differentiating the substitution provides a direct relationship that replaces differential terms.
• The integral often yields a simpler form, such as polynomials, which are easily integrated.

The given exercise demonstrates this by choosing a substitution that simplifies the polynomial structure, ultimately reducing the integral to a basic form. Successful application of integration techniques hinges on the efficiency and simplicity of the transformation process, aiding in finding solutions to otherwise challenging problems.

Variable Substitution

Variable substitution is a powerful tool in calculus for solving integrals, particularly when facing expressions that resist direct integration. This technique involves changing variables to express the integrand in a simpler form. The process includes:

• Identifying a substitution that turns complex terms into more tractable ones, often polynomials.
• Relating the differential of the chosen substitution directly to the integrand's differential.
• Changing limits of integration to correspond with the new variable if calculating a definite integral.

In this exercise, the substitution of variable \( u = t^2 - t \) transforms the integral into a manageable polynomial form, allowing straightforward integration.
This involves translating limits as well: original bounds on \( t \) become bounds on \( u \) post-substitution. Practicing variable substitution can significantly enhance problem-solving versatility in calculus, by breaking down otherwise intricate integrals into digestible parts.