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Chapter 11: Problem 61

Evaluate the following definite integrals. $$\int_{0}^{\ln 2}\left(e^{t} \mathbf{i}+e^{t} \cos \left(\pi e^{t}\right) \mathbf{j}\right) d t$$

Short Answer

Expert verified
Question: Evaluate the definite integral of the vector-valued function $$\int_{0}^{\ln 2}\left(e^{t} \mathbf{i}+e^{t} \cos \left(\pi e^{t}\right)\mathbf{j}\right) dt$$. Answer: $$\mathbf{i}$$

Step by step solution

01

Find the antiderivative of each component

To find the antiderivative of each component of the vector-valued function, we will integrate them separately: - Antiderivative of the first component (e^t * i): $$\int e^t dt = e^t + C_1$$ - Antiderivative of the second component (e^t * cos(pi * e^t) * j): $$\int e^t \cos(\pi e^t) dt$$ this is a bit trickier. We will use integration by substitution. Let u = pi * e^t, so du = pi * e^t dt. Therefore, dt = (1/pi) * (1/e^t) * du. Now, the integral becomes: $$\frac{1}{\pi} \int \cos(u) du = \frac{1}{\pi}( \sin(u) + C_2)$$ Now substitute back for u: $$\frac{1}{\pi}(\sin(\pi e^t) + C_2)$$
02

Evaluate the antiderivatives at the limits

Now we need to evaluate the antiderivatives at the limits, 0 and ln(2) and subtract the results. - Evaluate the antiderivative of the first component at the limits: $$e^{\ln 2} - e^0 = 2 - 1 = 1$$ So, the first component of the definite integral is equal to 1. - Evaluate the antiderivative of the second component at the limits: $$\frac{1}{\pi}\left(\sin\left(\pi e^{\ln 2}\right) - \sin\left(\pi e^0\right)\right) = \frac{1}{\pi}(\sin(\pi)- \sin(0))= \frac{1}{\pi} ( 0 - 0 )$$ So, the second component of the definite integral is equal to 0.
03

Write the result as a vector

Now, we just need to write the result as a vector. The definite integral of the given vector function is: $$\int_{0}^{\ln 2}\left(e^{t} \mathbf{i}+e^{t} \cos \left(\pi e^{t}\right)\mathbf{j}\right) dt = \left[ 1\mathbf{i} + 0\mathbf{j}\right] = \mathbf{i}$$

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Most popular questions from this chapter

Properties of dot products Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle,\) and \(\mathbf{w}=\left\langle w_{1}, w_{2}, w_{3}\right\rangle .\) Prove the following vector properties, where \(c\) is a scalar. $$\mathbf{u} \cdot(\mathbf{v}+\mathbf{w})=\mathbf{u} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{w}$$

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