Problem 45 - Calculus and its applications

Question

Evaluate. $$ \int_{0}^{1} x\left(x^{2}+1\right)^{5} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{21}{4} \).

Step 1: Identify the Integration Method

The given integral \[ \int_0^1 x(x^2+1)^5 \, dx \]suggests the use of substitution because it contains a composite function \((x^2 + 1)^5\). We notice that the derivative of \(x^2 + 1\) is \(2x\), which resembles part of the integrand.

Step 2: Choose a Substitution

Let \( u = x^2 + 1 \). Then, compute \( du \) as follows:\[ du = 2x \, dx \]This implies \( x \, dx = \frac{1}{2} \, du \). The limits of integration also change according to the substitution: when \( x = 0 \), \( u = 0^2 + 1 = 1 \); when \( x = 1 \), \( u = 1^2 + 1 = 2 \).

Step 3: Substitute and Simplify the Integral

Substitute \( u \) and \( du \) into the integral:\[ \int_{x=0}^{x=1} x(x^2+1)^5 \, dx = \int_{u=1}^{u=2} (u)^5 \cdot \frac{1}{2} \, du \]Simplify the integral:\[ \frac{1}{2} \int_{1}^{2} u^5 \, du \]

Step 4: Integrate Using the Power Rule

Integrate \( \frac{1}{2} \int_1^2 u^5 \, du \) using the power rule for integration. The power rule states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Thus:\[ \frac{1}{2} \cdot \left[ \frac{u^{6}}{6} \right]_1^2 \]Evaluate the definite integral by substituting the limits.

Step 5: Evaluate the Definite Integral

Substitute the upper and lower limits:\[ \frac{1}{2} \cdot \left( \frac{2^6}{6} - \frac{1^6}{6} \right) \]Calculate the values:\[ = \frac{1}{2} \cdot \left( \frac{64}{6} - \frac{1}{6} \right) \]\[ = \frac{1}{2} \cdot \frac{63}{6} = \frac{63}{12} = \frac{21}{4} \]

Step 6: Conclude the Solution

Thus, the evaluated integral of\[ \int_0^1 x(x^2+1)^5 \, dx \]is \( \frac{21}{4} \).

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus that simplifies the integration of composite functions. This technique involves choosing a substitution to transform the integral into an easier form to work with. In the given exercise, we identified that the function

• \( x(x^2+1)^5 \)

suggests using substitution because of the composite component \((x^2 + 1)^5\). By recognizing that the derivative of \(x^2+1\) is \(2x\), we can let \( u = x^2 + 1 \). This choice of \(u\) simplifies the integration process by transforming the original integrand into a function of \(u\), namely \( u^5 \). To complete the substitution, replace \(x \, dx\) in the integral with \(\frac{1}{2} \, du\), and adjust the limits of integration accordingly. When \(x=0\), \(u=1\), and when \(x=1\), \(u=2\). This results in a new integral:

• \( \int_{1}^{2} \frac{1}{2} u^5 \, du \)

which is much simpler to evaluate.
The substitution method essentially untangles complex functions, allowing for straightforward integration. With practice, selecting \(u\) becomes intuitive as you learn which substitutions simplify various forms.

Definite Integral

A definite integral represents the area under a curve within specified limits. It is written as:

• \( \int_{a}^{b} f(x) \, dx \)

and evaluated using fundamental calculus techniques. The result gives numerical value, representing the net area between the function and the x-axis over the interval \([a, b]\).
The exercise asked to evaluate the integral over the interval from \(x=0\) to \(x=1\). After applying substitution, the limits were transformed to correspond with the new variable \(u\). The new limits became \(u=1\) and \(u=2\). This means we need to determine the area under the curve between these new limits for \(u^5\).When evaluating a definite integral, it's crucial to substitute these limits back into the antiderivative obtained from the integration process. In this case:

• First, integrate \(u^5\) to get \(\frac{u^6}{6}\).
• Then substitute in the limits: \( \left[ \frac{u^6}{6} \right]_{1}^{2} \).

Finally, calculate the resulting expression. This gives us the accumulated area or value over that interval, providing insight into the behavior of the function within the specified range.

Power Rule for Integration

The power rule for integration is a fundamental technique used to integrate functions of the form \(u^n\). It is expressed as:

• \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \)

where \(n eq -1\) and \(C\) is the constant of integration, not needed when dealing with definite integrals. In our example, after substituting and transforming the integral, we ended up with:

• \( \frac{1}{2} \int_{1}^{2} u^5 \, du \)

Using the power rule, the integral of \(u^5\) becomes \(\frac{u^6}{6} \). This rule efficiently handles polynomials by increasing the exponent by one and dividing by the new exponent.
When calculating definite integrals, like in this problem, simply evaluate the antiderivative at the upper limit, subtract the value at the lower limit, and multiply by any constants outside the integrals. So in our steps:

• Compute \( \frac{1}{2} \cdot \left(\frac{2^6}{6} - \frac{1^6}{6}\right) \)

The power rule streamlines the integration of polynomial functions, turning potentially complex expressions into straightforward calculations.