Evaluate: quadint_(0)^(1)[e^(-t)i+(1)/(t+1)j]dt

Understand the integral: Understand the integral We are asked to evaluate the definite integral of a vector function from 0 to 1. The vector function is given by e^(-t)i + (1)/(t+1)j. This means we need to integrate each component of the vector function separately with respect to t over the interval [0, 1].Integrate the i component: Integrate the i component The i component of the vector function is e^(-t). We need to find the integral of e^(-t) from 0 to 1. The antiderivative of e^(-t) is -e^(-t). So we evaluate -e^(-t) from 0 to 1. int_(0)^(1)e^(-t)dt = [-e^(-t)]_(0)^(1) = -e^(-1) - (-e^(0)) = -e^(-1) + 1Integrate the j component: Integrate the j component The j component of the vector function is (1)/(t+1). We need to find the integral of (1)/(t+1) from 0 to 1. The antiderivative of (1)/(t+1) is ln|t+1|. So we evaluate ln|t+1| from 0 to 1. int_(0)^(1)(1)/(t+1)dt = [ln|t+1|]_(0)^(1) = ln|1+1| - ln|0+1| = ln(2) - ln(1) = ln(2)Combine the results: Combine the results Now we combine the results from Step 2 and Step 3 to get the vector result of the integral. The i component is -e^(-1) + 1 and the j component is ln(2). So the vector result is (-e^(-1) + 1)i + ln(2)j.

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Algebra 2

Sum of finite series not start from 1

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Q. Evaluate:

\int_{0}^{1} \left[ e^{-t} \mathbf{i} + \frac{1}{t+1} \mathbf{j} \right] dt

Understand the integral: Understand the integral $\newline$ We are asked to evaluate the definite integral of a vector function from $0$ to $1$ . The vector function is given by $e^{-t}i + \frac{1}{t+1}j$ . This means we need to integrate each component of the vector function separately with respect to $t$ over the interval $[0, 1]$ .
Integrate the i component: Integrate the i component $\newline$ The i component of the vector function is $e^{-t}$ . We need to find the integral of $e^{-t}$ from $0$ to $1$ . $\newline$ The antiderivative of $e^{-t}$ is $-e^{-t}$ . So we evaluate $-e^{-t}$ from $0$ to $1$ . $\newline$ $$\newline$\int_{$0$}^{$1$}e^{-t}dt = [-e^{-t}]_{$0$}^{$1$}$\newline$= -e^{$-1$} - (-e^{$0$})$\newline$= -e^{$-1$} + $1$
Integrate the j component: Integrate the j component$\newline$The j component of the vector function is $(1)/(t+1)$. We need to find the integral of $(1)/(t+1)$ from $0$ to $1$.$\newline$The antiderivative of $(1)/(t+1)$ is $\ln|t+1|$. So we evaluate $\ln|t+1|$ from $0$ to $1$.$\newline$$\int_{0}^{1}(1)/(t+1)\,dt = [\ln|t+1|]_{0}^{1}$$\newline$$(1)/(t+1)$$0$$\newline$$(1)/(t+1)$$1$$\newline$$(1)/(t+1)$$2$
Combine the results: Combine the results$\newline$Now we combine the results from Step $2$ and Step $3$ to get the vector result of the integral.$\newline$The $i$ component is $-e^{-1} + 1$ and the $j$ component is $\ln(2)$.$\newline$So the vector result is $(-e^{-1} + 1)i + \ln(2)j$.