5.3 The Fundamental Theorem of Calculus/9: Difference between revisions

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= 1\cdot((y)^{2}\sin{y})-0\cdot((0)^{2}\sin{(2)})
= 1\cdot((y)^{2}\sin{y})-0\cdot((0)^{2}\sin{(2)})
= y^{2}\sin{(y)}
= y^{2}\sin{(y)} \\


\textbf{Therefor, }  
\text{Therefore, } g'(y) = y^{2}\sin{(y)}


</math>
</math>

Revision as of 19:56, 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 \frac{d}{dy}\left[g(y)\right] = \frac{d}{dy}\left[\int_{2}^{y}t^2\sin{(t)}dt\right] = 1\cdot((y)^{2}\sin{y})-0\cdot((0)^{2}\sin{(2)}) = y^{2}\sin{(y)} \\ \text{Therefore, } g'(y) = y^{2}\sin{(y)} }