`int_1^2 (4x^3 - 3x^2 + 2x) dx` Evaluate the integral

Textbook Question

Chapter 5, 5.4 - Problem 22 - 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_1^2(4x^3 - 3x^2 + 2x)dx = int_1^2(4x^3)dx - int_1^2 3x^2 dx + int_1^2 2x dx`

Evaluating each definite integral, using the formula `int x^n dx = (x^(n+1))/(n+1) + c` , yields:

`int_1^2(4x^3)dx = 4*(x^4)/4|_1^2 = 2^4 - 1^4 = 15`

`int_1^2 3x^2 dx =3x^3/3|_1^2 = 2^3 - 1^3 = 7`

`int_1^2 2x dx = 2x^2/2|_1^2 = 2^2 - 1^2 = 3`

Gathering the results, yields:

`int_1^2(4x^3 - 3x^2 + 2x)dx = 15 - 7 + 3 = 11`

Hence, evaluating the definite integral, using the fundamental theorem of calculus yields `int_1^2(4x^3 - 3x^2 + 2x)dx = 11.`

scisser | (Level 3) Honors

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

`int_1^2(4x^3-3x^2+2x)dx=x^4-x^3+x^2|_1^2`

Plug in the upper limit and subtract away the lower limit

`[(2)^4-(2)^3+(2)^2]-[(1)^4-(1)^3+(1)^2]`

`=11`