`int_0^pi (5e^x + 3sin(x))dx` Evaluate the integral

Textbook Question

Chapter 5, 5.4 - Problem 27 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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You need to evaluate the definite integral using the fundamental theorem of calculus, such that: `int_a^b f(x)dx = F(b) - F(a)`

`int_0^pi (5e^x+ 3sin x)dx = int_0^pi 5e^x dx + int_0^pi 3sin x dx`

`int_0^pi (5e^x+ 3sin x)dx = (5e^x - 3cos x)|_0^pi`

`int_0^pi (5e^x+ 3sin x)dx =5e^pi - 3cos pi - 5e^0 + 3cos 0`

`int_0^pi (5e^x+ 3sin x)dx = 5e^pi - 3*(-1) - 5 + 3`

`int_0^pi (5e^x+ 3sin x)dx = 5e^pi + 1`

Hence, evaluating the definite integral yields

` int_0^pi (5e^x+ 3sin x)dx = 5e^pi + 1`

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