Question

Evaluate the integrals $$ \int_{-1}^{1}\left[(1+t)^{3 / 2} \mathbf{i}+(1-t)^{3 / 2} \mathbf{j}\right] d t $$

Short answer

\( \frac{8\sqrt{2}}{5} (\mathbf{i} + \mathbf{j}) \)

Step-by-step solution

Separate the Integral Components

The given integral is a vector integral and can be written in components: \[ \int_{-1}^{1} (1+t)^{3/2} \mathbf{i} \, dt + \int_{-1}^{1} (1-t)^{3/2} \mathbf{j} \, dt \] We will evaluate these two separate integrals one by one.

Evaluate the Integral of the i-component

Consider the first integral component: \[ \int_{-1}^{1} (1+t)^{3/2} \, dt \] To solve this integral, use the substitution method. Let \( u = 1+t \), then \( du = dt \). Change the limits of integration accordingly: - When \( t = -1 \), \( u = 0 \) - When \( t = 1 \), \( u = 2 \) The integral becomes: \[ \int_{0}^{2} u^{3/2} \, du \] Evaluate this integral using the power rule: \[ \frac{2}{5} u^{5/2} \Big|_0^2 = \frac{2}{5} (2^{5/2} - 0) = \frac{2}{5} \times 4\sqrt{2} = \frac{8\sqrt{2}}{5} \]

Evaluate the Integral of the j-component

Consider the second integral component: \[ \int_{-1}^{1} (1-t)^{3/2} \, dt \] Similar to the previous step, use the substitution \( v = 1-t \), then \( dv = -dt \). Change the limits of integration: - When \( t = -1 \), \( v = 2 \) - When \( t = 1 \), \( v = 0 \) The integral becomes: \[ -\int_{2}^{0} v^{3/2} \, dv = \int_{0}^{2} v^{3/2} \, dv \] Evaluate this integral using the power rule: \[ \frac{2}{5} v^{5/2} \Big|_0^2 = \frac{2}{5} (2^{5/2} - 0) = \frac{2}{5} \times 4\sqrt{2} = \frac{8\sqrt{2}}{5} \]

Combine Results

We have evaluated the two components separately: For the \( \mathbf{i} \) component: \( \frac{8\sqrt{2}}{5} \mathbf{i} \) For the \( \mathbf{j} \) component: \( \frac{8\sqrt{2}}{5} \mathbf{j} \) Combine these results to find the vector integral: \[ \frac{8\sqrt{2}}{5} \mathbf{i} + \frac{8\sqrt{2}}{5} \mathbf{j} = \frac{8\sqrt{2}}{5} (\mathbf{i} + \mathbf{j}) \]

Key concepts

Integral Calculus

Integral calculus is a branch of calculus focused on finding the integral of functions. Integrals are used to determine the area under curves, the total accumulated quantity, and much more. Simply put, while differential calculus is about the rate of change, integral calculus is about accumulation.

In the given problem, integral calculus is applied to a vector function, involving both magnitude and direction. This allows us to calculate not just scalar accumulations but also vector accumulations involving unit vectors like \(\mathbf{i}\) and \(\mathbf{j}\).

In integral calculus, the definite integral has a specific number range called limits of integration. For instance, \( \int_{-1}^{1} \) means that we accumulate the function's values from \(t = -1\) to \(t = 1\). This is crucial when calculating specific quantities and helps clearly define areas or totals between certain points.

Substitution Method

The substitution method is a technique used to simplify the process of integration. It involves substituting a part of the integral with a new variable to make the integral easier to solve. This transforms a potentially complex problem into a more straightforward one.

For example, in the given exercise, the substitution \( u = 1 + t \) was used. This substitution transforms the integral into a simpler form where the limits of integration and the differential \( dt \) are redefined in terms of \( u \). This kind of substitution aids in converting a complex function into a standard form where established rules like the power rule can be applied easily.

• Identify a part of the integrand that can be substituted.
• Change the limits of integration according to the new variable.
• Replace the integrand and differential with the new terms.

This step-by-step approach demystifies complicated integrals by turning them into basic standard forms.

Vector Integral

A vector integral is the integration process applied to vector-valued functions. Such functions include multiple components, each having its own integral.

In the exercise, a vector function \[ (1+t)^{3/2} \mathbf{i} + (1-t)^{3/2} \mathbf{j} \] needed to be integrated. To handle this, the overall vector integral was broken down into separate integrals for each component. This approach simplifies the integration, making it clear that vector integrals are just sums of simpler integrals.

• Decompose the vector function into its scalar components.
• Integrate each component separately.
• Recombining the results to form the final integrated vector.

Whether it's physics, engineering, or mathematics, vector integrals are particularly useful as they allow for computing curved path lengths, accumulated magnitudes, or even areas swept out by rotating bodies.

Power Rule

The power rule is a basic principle in calculus used to find the antiderivative of functions in the form \( x^n \). It states that the integral of \( x^n \) with respect to \( x \) is \( \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \).

In the problem, this rule was used after substitution to integrate terms like \( (1+t)^{3/2} \). The power rule makes it easy to compute the antiderivative of polynomials and other simple functions. Remember to adjust for any constants you factor out during integration to apply the rule quickly and effectively.

• Identify the exponent to apply the power rule.
• Write the antiderivative using \( \frac{x^{n+1}}{n+1} \).
• Evaluate the limits of integration using this antiderivative.

Mastering the power rule is essential for students because it simplifies many integration problems and provides a methodical approach to obtaining antiderivatives of straightforward polynomial expressions.