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I have a quick question. I need to find the derivative of $$ \int_{-x}^1\sqrt{t^2+1}\,dt$$

Why is the answer just $$\sqrt{x^2+1}\ ?$$ I know the 2nd FTC; but why is it not $$-\sqrt{x^2+1}\ ?$$

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    I assume you mean $dt$ not $dx$.2012-12-11

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We have$$F(x)=\int_{-x}^1 \sqrt{t^2+1}dx=-\int_{1}^{-x} \sqrt{t^2+1}dx$$ By the FTC and the chain rule, $$F^{\prime}(x)=-\sqrt{x^2+1}(-x)^{\prime}=\sqrt{x^2+1}$$

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    Your last line is incorrect. The derivative of $(-x)$ is $-1$ which cancels out with the other minus sign, just leaving you with $\sqrt{x^2+1}$.2012-12-12
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    @Joe Indeed. I fixed it2012-12-12
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Two cancelling minus signs. One comes from the fact the variable bound is in the lower position. The other is because the bound is $-x$, so we get an extra minus sign from the Chain Rule. The result is $-\left(-\sqrt{(-x)^2+1}\right)$, which is equivalent to the answer given, and not to your $-\sqrt{x^2+1}$.

Remark: If one wants to be formal, the integral is $F(0)-F(-x)$, where $F(t)$ is an antiderivative of $\sqrt{t^2+1}$. Now differentiate, using the fact that $F'(t)=\sqrt{t^2+1}$. One needs to remember to use the Chain Rule.