I am trying to learn calculus as part of a catch up course for a degree. Finding the derivative of a function makes sense to me but finding the integral of a function: I just can't get my head around.

This is the question: Find the indefinite integral and evaluate the definite integral from $-1$ to $+1$.

The function is written like this: $$f(x) = 2x^2 - 3x + 5$$ Could someone break down the method for integration of the above into small steps that I can wrap my head around? Thanks!


The indefinite integral of $ax^n$ is $\frac{a}{n+1}x^{n+1}$ and the integral of a sum is the sum of the integrals of the summands. Have you tried to sit down and work that out? For the definite integral from $A$ to $B$ you evaluate $F(B)-F(A)$, where $F$ is the primitive function (=the indefinite integral up to constants) of your function $f$.

I will add more details if you let me know where problems are arising for you.

  • $\begingroup$ Thanks for the answer! So the first bit makes sense it becomes $$\frac{2}{3}x^3$$ but for the $$3x+5$$ are you saying that I just add them together? so $$\frac{2}{3}x^3 + 8x + C$$ $\endgroup$ – Calum McManus Sep 25 '18 at 16:05
  • $\begingroup$ Oh wait, instead do you treat 3x the same as 3x^1? In that case it would be $$\frac{2}{3}x^3 + \frac{3}{2}x^2 + 5x + C$$ $\endgroup$ – Calum McManus Sep 25 '18 at 16:40
  • $\begingroup$ This is correct up to a minus sign before your second term. Now can you evaluate the definite integral? $\endgroup$ – user526015 Sep 25 '18 at 18:44
  • $\begingroup$ Yes that was a typo the first + should be a - :) I think so, what what I gather I write out the same equation 2 times without the constant and replace all the X's in each with the upper and then the lower limit and minus them from each other! $\endgroup$ – Calum McManus Sep 25 '18 at 19:03
  • $\begingroup$ That is what is to do, yes. But you keep the constants and they cancel out in the calculation. $\endgroup$ – user526015 Sep 25 '18 at 19:16

This question can be answered by the basic property of integral which states that an integral with a polynomial function will obey a distributive property. That is,

$$\int(f(x) + g(x))dx =\int f(x)dx + \int g(x)dx$$ so that \begin{align}\int(2x^2 – 3x + 5) dx & =\int 2x^2dx - \int 3x dx + \int5 dx \\ & = \frac{2x^3}{3} – \frac{3x^2}{2} + 5x + c\end{align}

  • $\begingroup$ So would that be $$2\frac{3x}{3} - 3\frac{2x}{2} + 5x + C$$ ? I am struggling to understand your formatting. $\endgroup$ – Calum McManus Sep 25 '18 at 17:25
  • $\begingroup$ No, the 3 after x is actually the power of x. And so on.. $\endgroup$ – ajanda Sep 25 '18 at 17:28
  • $\begingroup$ Ok so my other comment was correct? $$\frac{2}{3}x^3 - \frac{3}{2}x^2 + 5x + C$$ $\endgroup$ – Calum McManus Sep 25 '18 at 17:29
  • $\begingroup$ I am not able to write in the correct format $\endgroup$ – ajanda Sep 25 '18 at 17:29
  • 1
    $\begingroup$ You could avoid that kind of misunderstandings by using MathJax/LaTeX. $\endgroup$ – Henrik supports the community Sep 25 '18 at 17:31

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