<s>[INST] function - area under the curve Trying to find area under the curve for the function $f(x)= 4x^3 + 3x^2 - x + 1$ on the interval $[1,2]$.     
My answer is:                       
$$\int_{1}^2 4x^3 + 3x^2 - x + 1 dx = \left[\frac{4x^4}{4} + \frac{3x^3}{3} – \frac{x^2}{2} + x + c\right]_1^2$$
$$\left[\frac{4.2^4}{4} + \frac{3.2^3}{3} – 2 + 2\right] – \left[\frac{4.1}{4} + 1 – \frac{1}{2} + 1\right] = 24 – 2.5 = 21.5$$
Is this correct? Thanks
 [/INST] Yes this is correct. You have applied the fundamental theorem of calculus properly and computed the integral properly. Nice job! My only suggestion is to eliminate the bookkeeping of a constant $c$ in your result of the integration before the evaluation between bounds because this is usually the convention. So, for the second equality, you should have the equivalent expression (notice that the constant is irrelevant since it does not depend on the variable and thus is eliminated upon subtracting the evaluation at the lower bound from that at the upper bound)
$$\left[x^4 + x^3 - \frac{x^2}{2} + x\right]_1^2$$
in place of
$$\left[\frac{4x^4}{4} + \frac{3x^3}{3} - \frac{x^2}{2} + x + c\right]_1^2\,\,,$$
and then proceed to evaluate at the bounds as you have done here.
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