Okay I have to find the slope of this tangent to the curve $y=\int_0^x \frac{dx}{1+x^3}$ at the point where $x=1$.

My try- I integrated the expression and differentiated it afterwards to get the slope. But I'm not getting the correct answer on putting $x=1$ in the final expression that I'm getting after integration and diffrentiation.

Someone please help me out. Is there some other procedure to do questions like this?

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    $\begingroup$ Are you sure you don't mean $$y = \int_0^x \frac{dt}{1+t^3}$$ $\endgroup$ – ÍgjøgnumMeg Jul 9 '16 at 9:55
  • $\begingroup$ you want to calculate slope of the tangent than its of form $y=mx+c$ where m is the slope $\endgroup$ – Nebo Alex Jul 9 '16 at 9:56
  • $\begingroup$ Just an FYI it's bad practice to have the variable of integration as a limit on the integral. $\endgroup$ – okrzysik Jul 9 '16 at 9:56
  • $\begingroup$ @Ed_4434 yes I'm sure it's not that because then it'd be a case of leibnitz rule which its not $\endgroup$ – Zlatan Jul 9 '16 at 9:57
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    $\begingroup$ I'm simply pointing out what @okrzysik said; $t$ is a dummy variable. $\endgroup$ – ÍgjøgnumMeg Jul 9 '16 at 10:02

By the fundamental theorem of calculus:

$$\frac{dy}{dx} = \frac{1}{1+x^3} \Rightarrow \frac{dy}{dx}\bigg|_{x=1} = \frac{1}{1+1^3} = \frac{1}{2}$$


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