Question

Evaluate the following definite integrals. $${\int }_0^{\ln 2}(e^t\mathbf{i}+e^t\cos \left(\pi e^t\right)\mathbf{j})dt$$

Short answer

Question: Evaluate the definite integral of the vector-valued function $${\int }_0^{\ln 2}(e^t\mathbf{i}+e^t\cos \left(\pi e^t\right)\mathbf{j})dt$$. Answer: $$\mathbf{i}$$

Step-by-step solution

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: $$\frac{1}{\pi }(\sin (\pi e^t)+C_2)$$

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: $$\frac{1}{\pi }(\sin \left(\pi e^{\ln 2}\right)-\sin \left(\pi e^0\right))=\frac{1}{\pi }(\sin (\pi )-\sin (0))=\frac{1}{\pi }(0-0)$$ So, the second component of the definite integral is equal to 0.

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: $${\int }_0^{\ln 2}(e^t\mathbf{i}+e^t\cos \left(\pi e^t\right)\mathbf{j})dt=[1\mathbf{i}+0\mathbf{j}]=\mathbf{i}$$