Summers Kelley
09/02/2023 · Junior High School

\( \int _ { 0 } ^ { \ln 2 } \int _ { 0 } ^ { \ln 5 } e ^ { 2 x - y } d x d y \)

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To evaluate the double integral \[ \int_{0}^{\ln 2} \int_{0}^{\ln 5} e^{2x - y} \, dx \, dy, \] we can first compute the inner integral with respect to \(x\): \[ \int_{0}^{\ln 5} e^{2x - y} \, dx. \] Since \(e^{-y}\) is a constant with respect to \(x\), we can factor it out: \[ \int_{0}^{\ln 5} e^{2x - y} \, dx = e^{-y} \int_{0}^{\ln 5} e^{2x} \, dx. \] Next, we compute the integral \(\int_{0}^{\ln 5} e^{2x} \, dx\): \[ \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C. \] Now we evaluate this from \(0\) to \(\ln 5\): \[ \int_{0}^{\ln 5} e^{2x} \, dx = \left[ \frac{1}{2} e^{2x} \right]_{0}^{\ln 5} = \frac{1}{2} e^{2 \ln 5} - \frac{1}{2} e^{0} = \frac{1}{2} (5^2) - \frac{1}{2} (1) = \frac{25}{2} - \frac{1}{2} = \frac{24}{2} = 12. \] Thus, we have: \[ \int_{0}^{\ln 5} e^{2x - y} \, dx = e^{-y} \cdot 12 = 12 e^{-y}. \] Now we substitute this result back into the outer integral: \[ \int_{0}^{\ln 2} 12 e^{-y} \, dy. \] We can factor out the constant \(12\): \[ 12 \int_{0}^{\ln 2} e^{-y} \, dy. \] Now we compute the integral \(\int e^{-y} \, dy\): \[ \int e^{-y} \, dy = -e^{-y} + C. \] Evaluating this from \(0\) to \(\ln 2\): \[ \int_{0}^{\ln 2} e^{-y} \, dy = \left[-e^{-y}\right]_{0}^{\ln 2} = -e^{-\ln 2} - (-e^{0}) = -\frac{1}{2} + 1 = \frac{1}{2}. \] Thus, we have: \[ 12 \int_{0}^{\ln 2} e^{-y} \, dy = 12 \cdot \frac{1}{2} = 6. \] Therefore, the value of the double integral is \[ \boxed{6}. \]

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The value of the double integral is 6.
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