5.3 The Fundamental Theorem of Calculus/15: Difference between revisions

From Mr. V Wiki Math
Jump to navigation Jump to search
No edit summary
m (Protected "5.3 The Fundamental Theorem of Calculus/15" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite)))
 
(17 intermediate revisions by 2 users not shown)
Line 1: Line 1:
Use part 1 of the FTC to find the derivative of the function:
 
<math>y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt</math>
<math>y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt</math>


FTC 1:
<math>\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)\,dt=b^\prime{(x)}\cdot\,f(b(x))-\,a^\prime{(x)}\cdot\,f(a(x))</math>


<math>
<math>
\begin{align}
\begin{align}


\frac{d}{dx}
\frac{d}{dx}(y)= \frac{d}{dx}\left[\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\right]=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})
 
 
 
&= \int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt  
&= \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})-0\cdot\sqrt{0+\sqrt 0}
&= \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt tan(x)})


\end{align}
\end{align}
Line 20: Line 12:




 
<math>
 
\text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})
In this problem <math>a^\prime{(x)}= 0</math>, so when it is multiplied by <math>f(a(x))</math> it will result in 0 and doesn't need to be added.
</math>

Latest revision as of 20:15, 6 September 2022

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle y=\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt}


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{d}{dx}(y)= \frac{d}{dx}\left[\int_{0}^{tan(x)}\sqrt{t+\sqrt t}\,dt\right]=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}})-0\cdot\sqrt{0+\sqrt 0}\,=\sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}}) \end{align} }


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \text{Therefore, } y' = \sec^{2}(x)\cdot\sqrt{tan(x)+\sqrt{tan(x)}}) }

Retrieved from "https://wiki.dvaezazizi.com/index.php?title=5.3_The_Fundamental_Theorem_of_Calculus/15&oldid=3438"