`int_(-2)^0 ((1/2)t^4 + (1/4)t^3 - t) dt` Evaluate the integral

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Chapter 5, 5.4 - Problem 23 - 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_(-2)^0 ((1/2)t^4 + (1/4)t^3 - t)dt = int_(-2)^0 ((1/2)t^4 dt + int_(-2)^0 (1/4)t^3 dt - int_(-2)^0 tdt`

`int_(-2)^0 ((1/2)t^4 + (1/4)t^3 - t)dt = ((1/2)t^5/5 + (1/4)t^4/4 - t^2/2)|_(-2)^0`

`int_(-2)^0 ((1/2)t^4 + (1/4)t^3 - t)dt = ((1/2)0^5/5 + (1/4)0^4/4 - 0^2/2 -  (1/2)((-2)^5)/5- (1/4)((-2)^4)/4 + ((-2)^2)/2)`

`int_(-2)^0 ((1/2)t^4 + (1/4)t^3 - t)dt = 32/5 - 4 + 2`

`int_(-2)^0 ((1/2)t^4 + (1/4)t^3 - t)dt = 22/5`

Hence, evaluating the definite integral yields `int_(-2)^0 ((1/2)t^4 + (1/4)t^3 - t)dt = 22/5`

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