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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

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You can always verify with wolfram alpha: – Lucas Zanella Jun 8 '14 at 4:07
Thank you Lucas – user3052350 Jun 8 '14 at 4:11
Your approach and results are very correct. In order to verify, I suggest you first take the derivative of the antiderivative and check that the result is identical to the integrand. The remaining is simple. – Claude Leibovici Jun 8 '14 at 4:17

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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Yes, you are correct.

Suggestion: To calculate faster, you may write down the simplest form before substituting the numbers. e.g. change (4x^4)/4 to x^4.

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Thank you very much. – user3052350 Jun 8 '14 at 9:37

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