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

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<math>g(y)= \int_{y}^{2}t^2\sin{(t)}dt
<math>g(y)= \int_{2}^{y}t^2\sin{(t)}dt</math><br><br>


= </math>
<math>


\int_{0}^{1}\left(3+x\sqrt{x}\right)dx &= \int_{0}^{1}\left(3+x^{1}{x}^{\frac{1}{2}}\right)dx = \int_{0}^{1}\left(3+x^{1+\frac{1}{2}}\right)dx  = \int_{0}^{1}\left(3+x^{\frac{3}{2}}\right)dx \\[2ex]
\frac{d}{dy}\left[g(y)\right] = \frac{d}{dy}\left[\int_{2}^{y}t^2\sin{(t)}dt\right]


&= 3x+\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\bigg|_{0}^{1} = 3x+\frac{x^{\tfrac{5}{2}}}{\frac{5}{2}}\bigg|_{0}^{1} = 3x+\frac{2x^{\frac{5}{2}}}{5}\bigg|_{0}^{1} \\[2ex]
= 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>

Latest revision as of 20:14, 6 September 2022




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