Problem 34

Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int_{0}^{1} t(t-1)^{12} d t $$

Step-by-Step Solution

Verified

Answer

The integral evaluates to $-\frac{1}{182}$.

1Step 1: Choose Functions for Integration by Parts

In integration by parts, we use the formula $ \int u \, dv = uv - \int v \, du $. For the given integral, we let $ u = t $ and $ dv = (t-1)^{12} \, dt $. This choice makes $ du = dt $ and we will need to find $ v $ by integrating $ dv $.

2Step 2: Integrate dv to Find v

Find $ v $ by integrating $ dv = (t-1)^{12} \, dt $. The integral of $(t-1)^{12} $ is $ \frac{(t-1)^{13}}{13} $. Thus, $ v = \frac{(t-1)^{13}}{13} $.

3Step 3: Apply the Integration by Parts Formula

Substitute $ u = t $, $ du = dt $, $ dv = (t-1)^{12} \, dt $, and $ v = \frac{(t-1)^{13}}{13} $ into the integration by parts formula: $ \int_{0}^{1} t(t-1)^{12} \, dt = \left[ t \cdot \frac{(t-1)^{13}}{13} \right]_0^1 - \int_0^1 \frac{(t-1)^{13}}{13} \, dt $.

4Step 4: Evaluate the Boundary Terms

Evaluate the term $ \left[ t \cdot \frac{(t-1)^{13}}{13} \right]_0^1 $. When $ t = 1 $, this term is $ 1 \times \frac{0^{13}}{13} = 0 $. When $ t = 0 $, the term is $ 0 \times \frac{(-1)^{13}}{13} = 0 $. So, this entire boundary expression is $ 0 - 0 = 0 $.

5Step 5: Integrate Remaining Integral

Now integrate $ \int_0^1 \frac{(t-1)^{13}}{13} \, dt $. This becomes $ \frac{1}{13} \left[ \frac{(t-1)^{14}}{14} \right]_0^1 = \frac{1}{182} \left[ (0-1)^{14} - (1-1)^{14} \right] = \frac{1}{182} [1 - 0] = \frac{1}{182} $.

6Step 6: Final Answer

Combine the evaluated boundary terms and the integral computed to give the final answer: $ 0 - \frac{1}{182} = -\frac{1}{182} $.

Key Concepts

Definite IntegralCalculus TechniquesPolynomial Integration

Definite Integral

A definite integral is a type of integral evaluated over a specific interval, giving you a number that represents the accumulated area under a curve. It involves two limits, called the bounds of integration. For example, in \[ \int_0^1 t(t-1)^{12} \, dt \] the bounds are from 0 to 1. Understanding definite integrals is crucial because it helps us:

Determine the total accumulation of quantities - such as distance, area, or any measurable quantity over a specific interval.
Analyze real-world applications like calculating the total work done or finding the accumulated profit over time.

To compute a definite integral, follow these steps:

Find the antiderivative or the indefinite integral of the function. This involves standard calculus techniques like substitution or integration by parts.
Substitute the upper and lower bounds into the antiderivative.
Subtract the value of the antiderivative at the lower bound from its value at the upper bound.

This gives you the net area, or net change, over the interval.

Calculus Techniques

Calculus offers several techniques to solve integrals, especially when dealing with complex expressions. The method of integration by parts is one such technique and is especially useful when dealing with products of functions.The formula for integration by parts is \[ \int u \, dv = uv - \int v \, du \] and requires you to choose parts of the integral to substitute into the equation. For our example,

Choose $ u = t $ and $ dv = (t-1)^{12} \, dt $.
Then, differentiate $ u $ to find $ du $, and integrate $ dv $ to find $ v $.

This technique helps in breaking down the integral into simpler parts.When executed correctly, these techniques allow for evaluation of integrals that would be otherwise quite tricky. For example, by integrating $ dv = (t-1)^{12} \, dt $, we find $ v = \frac{(t-1)^{13}}{13} $, simplifying the integral evaluation.

Polynomial Integration

Polynomial integration involves integrating expressions that primarily include polynomials. This process, especially when dealing with high powers, relies on the power rule for integration.The power rule states that to integrate $ x^n $, where $ n eq -1 $, the antiderivative is \[ \frac{x^{n+1}}{n+1} \]. For our problem, the polynomial expression is $ (t-1)^{12} $. By applying the power rule, we find: