P. 244, line -11: Difference between revisions
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<math>\int_0^t \Xi^{-1}(\tau)g(\tau) d \tau = \int_0^t \left[ \begin{array}{c} -\frac{1}{2} \\ \frac{1}{2} e^{-2 \tau} \end{array} \right] d \tau = \left[ \begin{array}{c} -\frac{1}{2} t \\ \frac{1}{4} \left( 1 - e^{-2 t} \right) \end{array} \right].</math> | <math>\int_0^t \Xi^{-1}(\tau)g(\tau) d \tau = \int_0^t \left[ \begin{array}{c} -\frac{1}{2} \\ \frac{1}{2} e^{-2 \tau} \end{array} \right] d \tau = \left[ \begin{array}{c} -\frac{1}{2} t \\ \frac{1}{4} \left( 1 - e^{-2 t} \right) \end{array} \right].</math> | ||
[[Engineering Differential Equations: Theory and Applications, Springer 2010#Errata|Return to errata.]] | |||
![{\displaystyle \int _{0}^{t}\Xi ^{-1}(\tau )g(\tau )d\tau =\int _{0}^{t}\left[{\begin{array}{c}-{\frac {1}{2}}\\{\frac {1}{2}}e^{-2\tau }\end{array}}\right]d\tau =\left[{\begin{array}{c}-{\frac {1}{2}}t\\{\frac {1}{4}}\left(1-e^{-2t}\right)\end{array}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6efb516d7a5db744f002c3322b93dfd8fb08740d)
