Problem 56 - Calculus Single Variable

Question

Evaluate \(\int_{1}^{2} x(x-1)^{1 / 2} d x .\) For the first step, integrate by parts with \(u=x\)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{16}{15}\).

Step 1: Integration by Parts Setup

Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). Here, let \(u = x\) and \(dv = (x-1)^{1/2} \, dx\). This choice suggests setting \(du = dx\). To solve for \(v\), we must first address how to find \(v\) from \(dv\).

Step 2: Evaluate v

To find \(v\), we need to integrate \((x-1)^{1/2} \, dx\). Let \(w = x-1\), hence \(dw = dx\). The integral becomes \(\int w^{1/2} \, dw\). Using the power rule for integration, \(\int w^n \, dw = \frac{w^{n+1}}{n+1} + C\), we get \(v = \frac{2}{3}(x-1)^{3/2}\).

Step 3: Apply Integration by Parts Formula

Substituting back into the integration by parts formula, we have \(uv = x \cdot \frac{2}{3}(x-1)^{3/2}\) and \(\int v \, du = \int \frac{2}{3}(x-1)^{3/2} \, dx\). The integral becomes: \(\left[ x \cdot \frac{2}{3}(x-1)^{3/2} \right]_1^2 - \int_1^2 \frac{2}{3}(x-1)^{3/2} \, dx\).

Step 4: Evaluate the Boundary Conditions

For the term \(uv\), substitute the upper and lower limits to get: \(\left[ \frac{2}{3} \cdot 2 \cdot 1^{3/2} - \frac{2}{3} \cdot 1 \cdot 0^{3/2} \right] = \frac{4}{3}\). For the integral: \(\int \frac{2}{3}(x-1)^{3/2} \, dx\), use substitution by evaluating: \[\left[ \frac{2}{3} \cdot \frac{2}{5} (x-1)^{5/2} \right]_1^2\].

Step 5: Evaluate the New Integral

Calculate the new integral evaluated from 1 to 2: \(\left[ \frac{4}{15} (x-1)^{5/2} \right]_1^2 = \frac{4}{15} \cdot (1^{5/2} - 0^{5/2}) = \frac{4}{15}\).

Step 6: Combine Results

Combine the results from step 4 and 5: \[\frac{4}{3} - \frac{4}{15} = \frac{20}{15} - \frac{4}{15} = \frac{16}{15}\]. Thus, the value of the integral is \(\frac{16}{15}\).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus. They calculate the area under a curve between two points, known as boundaries. When you see an expression like \( \int_{a}^{b} f(x) \, dx \), it represents the definite integral of a function \( f(x) \) from boundary \( a \) to \( b \).

This tells us how much "area" the function covers from one point to another. This area is always with respect to the x-axis. Here, we are dealing with \( \int_{1}^{2} x(x-1)^{1/2} \, dx \). The \( 1 \) and \( 2 \) are your limits of integration. They tell you the range over which you need to find the area.

For this exercise, you'll compute two main parts: evaluating the integrated function at the upper boundary minus evaluating it at the lower boundary.

Power Rule for Integration

The power rule for integration is a handy tool when dealing with polynomials or power expressions. It simplifies finding the integral of expressions like \( x^n \), where \( n \) is a real number. The rule says:

• The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the integration constant.
• Make sure \( n eq -1 \).

This rule is used in evaluating the integral of \( (x-1)^{1/2} \) in our exercise. Here, substitute \( w = x-1 \), making the integration look like \( \int w^{1/2} \, dw \).

Using the power rule, this becomes \( \frac{w^{3/2}}{3/2} + C \), simplifying to \( \frac{2}{3} w^{3/2} \). To help understand, try applying the rule to different powers and observe how it changes the integral.

Integration Techniques

Sometimes, directly integrating a function isn't straightforward. This is where different integration techniques come into play. They help break down complex integrals into manageable parts. In our exercise, we employ Integration by Parts.

Integration by Parts comes from the product rule for differentiation and is expressed as:

• \( \int u \, dv = uv - \int v \, du \).

Here, an intelligent choice of \( u \) and \( dv \) can simplify the process. Choosing \( u = x \) and \( dv = (x-1)^{1/2} \, dx \) allows us to separate the integral into simpler parts. This technique is especially useful when you have a product of functions.

It ensures that you can handle tough integrations by transforming them into something more workable.

Boundary Evaluation

Boundary evaluation is crucial in computing definite integrals. Once integration is performed, the next step is evaluating the primitive function at the boundaries given. For our integral, the bounds are 1 and 2.

This means you compute the integrated expression at \( x = 2 \) and \( x = 1 \) and subtract the results. This process can be summarized as:

• Substitute the upper bound into the integrated function.
• Subtract the result when you substitute the lower bound.

For example, when you compute \( [ x \cdot \frac{2}{3}(x-1)^{3/2} ]_1^2 \), you02 calculate the expression for both bounds and then find their difference.

Understanding boundary evaluation helps you find the exact value representing the area under a curve, which is crucial for determining the correctness of a definite integral.