find the area bounded by the curves $y = x^2$ and $y = 5x-6$.
First method I tried was to find where the two curves meet, which is at $x = 2, x =3$.
Then integrated $x^2$ with limits from $0$ to $2$ and subtracted the integral of $5x-6$ withs limits from $0$ to $2$ got $4.67$ as the answer but was marked wrong.
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$\begingroup$It's just $\int\limits_2^3(5x-6-x^2)dx=\left.\frac{5x^2}{2}-6x-\frac{x^3}{3}\right|_2^3=\frac{1}{6}$
$\endgroup$ 3 $\begingroup$You know the intersection of the curves is at $x = 2, x = 3$, so simply find which curve is "above" the other in the y-axis within these bounds (it's $y = 5x - 6$) and subtract the other function ($y = x^2$). So your final integral is $\int_2^3 5x - 6 - x^2 dx = \frac{5}{2}x^2 - 6x - \frac{1}{3}x^3|_2^3 = 1/6$. You just made a mistake in choosing your bounds.
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