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The area between $y = x^2$ and $y = x$ from $x = 0$ to $x = 1$ is:
A$1/6$, since $\int_0^1 (x - x^2)\, dx = 1/2 - 1/3 = 1/6$
B$1/3$, the integral of $x^2$ alone over the interval $[0, 1]$
C$1/2$, the integral of $x$ alone on the interval $[0, 1]$
D$5/6$, the sum of the two integrals on the interval $[0, 1]$
Answer & Solution
Correct answer: A. $1/6$, since $\int_0^1 (x - x^2)\, dx = 1/2 - 1/3 = 1/6$
$\int_0^1 (x - x^2)\, dx = 1/2 - 1/3 = 1/6$.
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