`int_(-2)^3 (x^3 - 3)dx` Evaluate the integral

Textbook Question

Chapter 5, 5.4 - Problem 21 - 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 integral, hence, you need to use the fundamental theorem of calculus, such that:

`int_a^b f(x)dx = F(b) - F(a)`

`int_(-2)^3(x^3 - 3)dx = int_(-2)^3(x^3)dx - int_(-2)^3 3 dx`

Evaluating each definite integral, using the formula ` int_a^b x^n dx = (x^(n+1))/(n+1)|_a^b` yields:

`int_(-2)^3(x^3)dx = (x^4)/4|_(-2)^3 = (1/4)(3^4 - (-2)^4) = 65/4`

`int_(-2)^3 3 dx =3x|_(-2)^3 = 3(3 - (-2)) = 15`

`int_(-2)^3(x^3 - 3)dx = 65/4 - 15 = 5/4`

Hence, evaluating the definite integral, using the fundamental theorem of calculus yields `int_(-2)^3(x^3 - 3)dx = 5/4.`

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scisser | (Level 3) Honors

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Find the antiderivative of each term: `int (a^n=(a^(n+1))/(n+1))`

`int_-2^3(x^3-3)=x^4/4-3x|_-2^3`

 

Plug in the upper limit and subtract away the lower limit

`[(3)^4/4-3(3)]-[(-2)^4/4-3(-2)]=1.25 or 5/4`

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