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Q. Evaluate:
- Understand the integral: Understand the integralWe are asked to evaluate the definite integral of a vector function from to . The vector function is given by . This means we need to integrate each component of the vector function separately with respect to over the interval .
- Integrate the i component: Integrate the i componentThe i component of the vector function is . We need to find the integral of from to .The antiderivative of is . So we evaluate from to .$\(\newline\)\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 \((-e^{-1} + 1)i + \ln(2)j\).
