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Chapter 10: Problem 47

Find the indefinite integral. $$ \int\left[(2 t-1) \mathbf{i}+4 t^{3} \mathbf{j}+3 \sqrt{t} \mathbf{k}\right] d t $$

Short Answer

Expert verified
The indefinite integral of the given vector is \((t^2 - t) \mathbf{i} + t^{4} \mathbf{j} + 2t^{3/2} \mathbf{k} + C\).

Step by step solution

01

Integrate the components of the Vector i

Start off by solving the integral for the i-component of the vector, which is \(2t - 1\). The indefinite integral of this is defined by \( \int (2t - 1) dt\). The rule of integrating any term \(at^n\), where \(a\), \(n\) are constants and \(t\) is the variable to be integrated is given by \((a/(n+1))*t^{n+1}\). By applying this rule, \(\int (2t - 1) dt = t^2 - t\).
02

Integrate the components of the Vector j

The j-component of our vector is given by \(4t^3\). By applying the same rule of integral mentioned in Step 1, \(\int 4t^3 dt\) is found to be \(t^4\).
03

Integrate the components of the Vector k

The k-component is given by \(3\sqrt{t} = 3t^{1/2}\). The indefinite integral \(\int 3t^{1/2} dt\) can be computed using the rule of integral used in the previous steps and is found to be \(2t^{3/2}\).
04

Combine the Results

We have found the indefinite integrals of the i, j and k vectors independently. Now, combine these results for the final solution: \[\int\left[(2 t-1) \mathbf{i}+4 t^{3} \mathbf{j}+3 \sqrt{t} \mathbf{k}\right] dt = (t^2 - t) \mathbf{i} + t^{4} \mathbf{j} + 2t^{3/2}\mathbf{k}\] + C, where C is the constant of integration which arises in indefinite integrals.

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Most popular questions from this chapter

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