How do you find the definite integral of $\int (x^{2} - x) d x$ $int (x^2-x)dx$ from $[0, 2]$ $[0,2]$ ?

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How do you find the definite integral of $\int (x^{2} - x) d x$ $int (x^2-x)dx$ from $[0, 2]$ $[0,2]$ ?

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Answer 1 · 2015-11-21T16:18:36

$\int_{0}^{2} (x^{2} - x) d x = \frac{2}{3}$ $int_0^2(x^2-x)dx = 2/3$

Explanation:

We have

$\int_{0}^{2} (x^{2} - x) d x$ $int_0^2(x^2 - x)dx$

Since it's a sum we can break the integrals into two parts

$\int_{0}^{2} (x^{2} - x) d x = \int_{0}^{2} x^{2} d x + \int_{0}^{2} - x d x$ $int_0^2(x^2-x)dx = int_0^2x^2dx + int_0^2-xdx$

We can integrate these two easily

$\int_{0}^{2} (x^{2} - x) d x = \frac{x^{3}}{3} {∣ ∣ ∣_0^{2} - \frac{x^{2}}{2} ∣ ∣ ∣}_{0}^{2}$ $int_0^2(x^2-x)dx = x^3/3|_0^2 - x^2/2|_0^2$

Or,

$\int_{0}^{2} (x^{2} - x) d x = \frac{2^{3}}{3} - \frac{0^{3}}{3} - \frac{2^{2}}{2} + \frac{0^{2}}{2} = \frac{8}{3} - 2 = \frac{8 - 6}{3} = \frac{2}{3}$ $int_0^2(x^2-x)dx = 2^3/3 - 0^3/3 - 2^2/2 + 0^2/2 = 8/3 - 2 = (8-6)/3 = 2/3$

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How do you find the definite integral of $\int (x^{2} - x) d x$ $int (x^2-x)dx$ from $[0, 2]$ $[0,2]$ ?

1 Answer

Explanation:

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How do you find the definite integral of ∫(x2−x)dxint (x^2-x)dx from [0,2][0,2]?

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How do you find the definite integral of $\int (x^{2} - x) d x$ $int (x^2-x)dx$ from $[0, 2]$ $[0,2]$ ?