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Section 5.7 : Computing Definite Integrals
13. Evaluate the following integral, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.
\[\int_{0}^{2}{{{{\bf{e}}^x} + \frac{1}{{{x^2} + 1}}\,dx}}\]Show All Steps Hide All Steps
Start SolutionFirst we need to integrate the function.
\[\int_{0}^{2}{{{{\bf{e}}^x} + \frac{1}{{{x^2} + 1}}\,dx}} = \left. {\left( {{{\bf{e}}^x} + {{\tan }^{ - 1}}\left( x \right)} \right)} \right|_0^2\]Recall that we don’t need to add the “+c” in the definite integral case as it will just cancel in the next step.
Show Step 2The final step is then just to do the evaluation.
We’ll leave the basic arithmetic to you to verify and only show the results of the evaluation. Make sure that you evaluate the upper limit first and then subtract off the evaluation at the lower limit.
Here is the answer for this problem.
\[\int_{0}^{2}{{{{\bf{e}}^x} + \frac{1}{{{x^2} + 1}}\,dx}} = \left( {{{\bf{e}}^2} + {{\tan }^{ - 1}}\left( 2 \right)} \right) - \left( {{{\bf{e}}^0} + {{\tan }^{ - 1}}\left( 0 \right)} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{{{\bf{e}}^2} + {{\tan }^{ - 1}}\left( 2 \right) - 1}}\]Note that \({\tan ^{ - 1}}\left( 0 \right) = 0\) but \({\tan ^{ - 1}}\left( 2 \right)\) doesn’t have a “nice” answer and so was left as is.