`int_0^4 (3sqrt(t) - 2e^t) dt` Evaluate the integral

Textbook Question

Chapter 5, 5.4 - Problem 30 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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_0^4(3sqrt t - 2e^t)dt =  int_0^4 3sqrt tdt - int_0^4 2e^t dt`

`  int_0^4(3sqrt t - 2e^t)dt = (3(t^(1/2+1))/(1/2+1) - 2e^t)|_0^4`

 

`int_0^4(3sqrt t - 2e^t)dt = (3(4^(1/2+1))/(1/2+1) - 2e^4+ 2e^0)`

`int_0^4(3sqrt t - 2e^t)dt = (3(2/3)*8 - 2e^4+ 2)`

`int_0^4(3sqrt t - 2e^t)dt = (18 - 2e^4)`

Hence, evaluating the definite integral, yields `int_0^4(3sqrt t - 2e^t)dt = (18 - 2e^4).`

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