5.3 The Fundamental Theorem of Calculus/8: Difference between revisions
No edit summary |
No edit summary |
||
| Line 2: | Line 2: | ||
\begin{align} | \begin{align} | ||
g(x)=\int_{3}^{x}e^{t^2-t}dt \\ | g(x)=\int_{3}^{x}e^{t^2-t}dt \\ | ||
\frac{d}{dx}\left[g(x)\right] = \frac{d}{dx}\left[\int_{3}^{x}e^{t^2-t}dt\right]=1e^{x^2-x}-0e^{3^2-3}=e^{x^2-x} | \frac{d}{dx}\left[g(x)\right] = \frac{d}{dx}\left[\int_{3}^{x}e^{t^2-t}dt\right]=1e^{x^2-x}-0e^{3^2-3}=e^{x^2-x} \\ | ||
\text{Therefore, } g'(x)=e^{x^2-x} | \text{Therefore, } g'(x)=e^{x^2-x} | ||
\end{align} | \end{align} | ||
Revision as of 20:33, 23 August 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 \begin{align} g(x)=\int_{3}^{x}e^{t^2-t}dt \\ \frac{d}{dx}\left[g(x)\right] = \frac{d}{dx}\left[\int_{3}^{x}e^{t^2-t}dt\right]=1e^{x^2-x}-0e^{3^2-3}=e^{x^2-x} \\ \text{Therefore, } g'(x)=e^{x^2-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 \begin{align} u & = \tfrac{1}{\sqrt{2}}(x+y) \qquad & x &= \tfrac{1}{\sqrt{2}}(u+v) \\[0.6ex] v & = \tfrac{1}{\sqrt{2}}(x-y) \qquad & y &= \tfrac{1}{\sqrt{2}}(u-v) \end{align}}