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Given the problem find $F'(x)$:

$$ \int_x^{x+2} (4t+1) \ \mathrm{dt}$$

I just feel stuck and don't know where to go with this, we learned the second fundamental theorem of calculus today but i don't know where to plug it in. What i did:

  • chain rule doesn't really take into effect here(*1) so just replace t with $x$
  • $F'(x) = 4x + 1$

though the answer is just 8, what am i doing wrong?

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$\int_x^{x+2} = \int_x^0 + \int_0^{x+2}$ here, and $\int_x^0 = - \int_0^x$. –  copper.hat Dec 13 '12 at 0:02
You didn't tell us what $F$ is. –  David Mitra Dec 13 '12 at 0:13
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2 Answers

up vote 6 down vote accepted

Let $g(t)=4t+1$, and let $G(t)$ be an antiderivative of $g(t)$. Note that $$F(x)=G(x+2)-G(x).\tag{$1$}$$

In this case, we could easily find $G(t)$. But let's not, let's differentiate $F(x)$ immediately. Since $G'(t)=g(t)=4t+1$. we get $$F'(x)=g(x+2)-g(x)=[4(x+2)+1]-[4x+2].$$ This right-hand side ismplifies to $8$.

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thanks! question makes a whole lot more sense now. –  Need4Sleep Dec 13 '12 at 0:11
In this case, you could easily have integrated $4t+1$. But you will be asked similar questions where $g(t)$ is difficult or impossible to integrate. The only other "twist" is when you are dealing with say $\int_0^{x^2} g(t)\,dt$. When you differentiate $G(t^2)-G(0)$, you need to use the Chain Rule. –  André Nicolas Dec 13 '12 at 0:15
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The second fundamental theorem lets you differentiate $\int_k^x f(t)\,dt$ with respect to $x$ where $k$ is a constant. Note that there's only one $x$ in the limits of the integral here.

Can you rewrite your integral as a difference of two integrals, each with just one $x$ in the limits of integration? Then you can try to apply the second fundamental theorem to each one.

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would x to x+2 be the same as 0 to 2? considering the difference of x to x+2 is 2? –  Need4Sleep Dec 12 '12 at 23:57
No, integrals don't usually work that way. Do you know how to simplify $\int_a^b f(t)\,dt + \int_b^c f(t)\,dt$? And what happens to an integral if you swap the limits of integration? –  user108903 Dec 12 '12 at 23:58
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