Math  /  Calculus

Question3. Evaluate the following definite integrals (a) 01(e2xexex)dx[5pts]\int_{0}^{1}\left(\frac{e^{2 x}-e^{-x}}{e^{x}}\right) d x[5 \mathrm{pts}]

Studdy Solution
Evaluate the definite integral from 0 to 1.
Substitute the limits into the integrated function:
[ex+12e2x]01\left[ e^{x} + \frac{1}{2} e^{-2x} \right]_{0}^{1}
Calculate at the upper limit x=1x = 1:
e1+12e2×1=e+12e2e^{1} + \frac{1}{2} e^{-2 \times 1} = e + \frac{1}{2} e^{-2}
Calculate at the lower limit x=0x = 0:
e0+12e2×0=1+12×1=1+12e^{0} + \frac{1}{2} e^{-2 \times 0} = 1 + \frac{1}{2} \times 1 = 1 + \frac{1}{2}
Subtract the lower limit result from the upper limit result:
(e+12e2)(1+12)\left( e + \frac{1}{2} e^{-2} \right) - \left( 1 + \frac{1}{2} \right)
Simplify:
e+12e2112e + \frac{1}{2} e^{-2} - 1 - \frac{1}{2}
Combine like terms:
e1+12(e21)e - 1 + \frac{1}{2} (e^{-2} - 1)
This is the value of the definite integral:
e32+12e2\boxed{e - \frac{3}{2} + \frac{1}{2} e^{-2}}

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