Chapter 12: Problem 21
In Exercises cvaluate the integral. $$ \int_{-1}^{1}\left[(1+t)^{3 / 2} \mathbf{i}+(1-t)^{3 / 2} \mathbf{j}\right] d t $$
Short Answer
Expert verified
The integral evaluates to \( \frac{8\sqrt{2}}{5} \mathbf{i} + \frac{8\sqrt{2}}{5} \mathbf{j} \).
Step by step solution
01
Write the Integral in Scalar Form
The given integral is a vector integral and can be written in terms of its components. We have two parts: \( \int_{-1}^{1} (1+t)^{3/2} \mathbf{i} \,dt \) and \( \int_{-1}^{1} (1-t)^{3/2} \mathbf{j} \,dt \). Each is a separate scalar integral that we can solve independently.
02
Evaluate the First Component
Begin with the integral \( \int_{-1}^{1} (1+t)^{3/2} \, dt \). Make a substitution; let \( u = 1 + t \), then \( du = dt \) and when \( t = -1 \), \( u = 0 \) and when \( t = 1 \), \( u = 2 \). This transforms the integral to \( \int_{0}^{2} u^{3/2} \, du \).
03
Solve the First Component Integral
The integral \( \int u^{3/2} \, du \) can be evaluated as \( \frac{2}{5}u^{5/2} \). Therefore, \[ \int_{0}^{2} u^{3/2} \, du = \left[ \frac{2}{5} u^{5/2} \right]_{0}^{2} = \frac{2}{5}(2^{5/2}) - \frac{2}{5}(0^{5/2}) = \frac{2}{5} \times 4\sqrt{2} = \frac{8\sqrt{2}}{5}. \]
04
Evaluate the Second Component
Next, evaluate the integral \( \int_{-1}^{1} (1-t)^{3/2} \, dt \). Use a similar substitution; let \( v = 1-t \), then \( dv = -dt \), and the limits change from \( t = -1 \) to \( v = 2 \) and \( t = 1 \) to \( v = 0 \). This transforms into \( -\int_{2}^{0} v^{3/2} \, dv = \int_{0}^{2} v^{3/2} \, dv \).
05
Solve the Second Component Integral
The integral \( \int v^{3/2} \, dv \) is solved similarly as \( \frac{2}{5} v^{5/2} \). Hence, \[ \int_{0}^{2} v^{3/2} \, dv = \left[ \frac{2}{5} v^{5/2} \right]_{0}^{2} = \frac{2}{5}(2^{5/2}) - \frac{2}{5}(0^{5/2}) = \frac{2}{5} \times 4\sqrt{2} = \frac{8\sqrt{2}}{5}. \]
06
Combine the Results
Finally, combine the results of the two integrals to get the vector result: \( \left( \frac{8\sqrt{2}}{5} \right)\mathbf{i} + \left( \frac{8\sqrt{2}}{5} \right)\mathbf{j} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Integral
Vector integrals are an extension of regular integrals where each component of a vector function is integrated separately. They are particularly useful in fields such as physics and engineering, where vector fields describe phenomena like electromagnetic fields or fluid flow. For a vector integral like \[\int \mathbf{F}(t) \, dt = \int P(t)\mathbf{i} + Q(t)\mathbf{j} \, dt,\] you need to evaluate each component independently:
- For the function \[P(t)\mathbf{i},\]the integral becomes \[\int P(t)\, dt,\]
- For the function \[Q(t)\mathbf{j},\]the integral becomes \[\int Q(t)\, dt.\]
Substitution Method
The substitution method is a powerful technique for solving integrals, particularly when the integrand contains a composite function. This method changes the variable of integration to simplify the integral. In the context of the exercise provided, the substitution plays a crucial role in handling trigonometric and polynomial forms:
- Choose a substitution: Such as letting \(u = 1 + t\), transforms \(dt\) to \(du\).
- Adjust the limits of integration: If the original integral is from \(-1\) to \(1\), you recalculate these limits in terms of the new variable \(u\).
- Transform the integral completely: The integral \(\int_{-1}^{1} (1+t)^{3/2} \, dt\) becomes \(\int_{0}^{2} u^{3/2} \, du\).
- Evaluate the new integral and reverse the substitution if necessary.
Scalar Integrals
Scalar integrals refer to the integration of scalar functions, which are functions that output a single value, rather than a vector. In vector calculus, sometimes you break down a vector integral into multiple scalar integrals. This simplification process happens when considering each component of a vector separately.
- For instance, in the given problem, \[\int (1+t)^{3/2}\,dt\]is a scalar integral representing part of a larger vector integral.
- To solve, using a standard formula or technique, such as power rule or substitution, might be applied.