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### Section 5-1 : Indefinite Integrals

2. Evaluate each of the following indefinite integrals.

1. $$\displaystyle \int{{40{x^3} + 12{x^2} - 9x + 14\,dx}}$$
2. $$\displaystyle \int{{40{x^3} + 12{x^2} - 9x\,dx}} + 14$$
3. $$\displaystyle \int{{40{x^3} + 12{x^2}\,dx}} - 9x + 14$$
Hint : As long as you recall your derivative rules and the fact that all this problem is really asking is the for us to determine the function that we differentiated to get the integrand (i.e. the function inside the integral….) this problem shouldn’t be too difficult.

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a $$\displaystyle \int{{40{x^3} + 12{x^2} - 9x + 14\,dx}}$$ Show Solution

All we are being asked to do here is “undo” a differentiation and if you recall the basic differentiation rules for polynomials this shouldn’t be too difficult. As we saw in the notes for this section all we really need to do is increase the exponent by one (so upon differentiation we get the correct exponent) and then fix up the coefficient to make sure that we will get the correct coefficient upon differentiation.

Here is that answer for this part.

$\int{{40{x^3} + 12{x^2} - 9x + 14\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{10{x^4} + 4{x^3} - \frac{9}{2}{x^2} + 14\,x + c}}$

Don’t forget the “+c”! Remember that the original function may have had a constant on it and the “+c” is there to remind us of that.

Also, don’t forget that you can easily check your answer by differentiating your answer and making sure that the result is the same as the integrand.

b $$\displaystyle \int{{40{x^3} + 12{x^2} - 9x\,dx}} + 14$$ Show Solution

This part is not really all that different from the first part. The only difference is the placement of the dx. Recall that one of the things that the dx tells us where to end the integration. So, in the part we are only going to integrate the first term.

Here is the answer for this part.

$\int{{40{x^3} + 12{x^2} - 9x\,dx}} + 14 = \require{bbox} \bbox[2pt,border:1px solid black]{{10{x^4} + 4{x^3} - \frac{9}{2}{x^2}\, + c + 14}}$

c $$\displaystyle \int{{40{x^3} + 12{x^2}\,dx}} - 9x + 14$$ Show Solution

The only difference between this part and the previous part is that the location of the dx moved.

Here is the answer for this part.

$\int{{40{x^3} + 12{x^2}\,dx}} - 9x + 14 = \require{bbox} \bbox[2pt,border:1px solid black]{{10{x^4} + 4{x^3} + c - 9x + 14}}$