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The first part of Fundamental Theorem of Calculus gives an abstract way of producing antiderivatives:

Write $F(x) = \int^x_1 (t + 1)dt$ in an explicit form. Check that its derivative is indeed $x + 1$.

What does it mean to find an integral in it explicit form? Does it mean to find its anti derivative?

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  • $\begingroup$ The second portion, "Write F(x) ... is indeed x+1.", is very hard to understand. Please consider writing it in very clear LaTeX. $\endgroup$ – Arturo don Juan Jan 15 '15 at 4:32
  • $\begingroup$ I disagree; it was at worst mildly annoying to read. I edited it anyway, because the (calculus) tag was appropriate. $\endgroup$ – Eric Stucky Jan 15 '15 at 4:37
  • $\begingroup$ @Kaaagome: However, you have made over 30 contributions to this site. You should consider learning the basics of LaTeX; at the very worst, you will still be able to write gorgeous equations. $\endgroup$ – Eric Stucky Jan 15 '15 at 4:41
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Here you have a "definite integral" with a variable in one of the limits, so you need to find the antiderivative and actually substitute the limits into it. The general antiderivative is $F(t)=t^2/2+t+C$ so when you substitute the limits you get $x^2/2+x-3/2$.

Edit: Since the general antiderivative is $F(t)=t^2/2+t+C$, you find your integral by substituting in the limits. That is,

$$\int_1^x (t+1)dt = F(x)-F(1)=x^2/2+x+C-(3/2+C)=x^2/2+x-3/2.$$

This is not really any different than you would do if the upper limit were $2$ or $4/3$, except that the result depends on the variable $x$.

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  • $\begingroup$ I get how you found the general anti derivative, but what od you mean substitute the limits into it? $\endgroup$ – Elsa Jan 15 '15 at 4:52
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    $\begingroup$ @Kaaagome See edit. Also I don't entirely understand the downvotes here. $\endgroup$ – Ian Jan 15 '15 at 4:55
  • $\begingroup$ Thanks for the edit! So basically the explicit form is just finding the anti derivative? $\endgroup$ – Elsa Jan 15 '15 at 5:30
  • $\begingroup$ @Kaaagome More or less, except for the $-3/2$ from the lower limit of $1$. Anyway, the point of the problem is just checking the FTC in this special case. $\endgroup$ – Ian Jan 15 '15 at 6:22

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