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

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<math>g(y)= \int_{2}^{y}t^2\sin{(t)}dt</math><br>
<math>g(y)= \int_{2}^{y}t^2\sin{(t)}dt</math><br><br>
= 1\cdot(y^{2}sin{(y)})-0\cdot(2^{2}sin{(2)})<br>
 
= y^{2} sin{(y)}  
<math>
 
\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)}
 
</math>
 
 
<math>
\text{Therefore, } g'(y) = y^{2}\sin{(y)}
</math>
</math>

Latest revision as of 20:14, 6 September 2022