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Question Number 121775 by ZiYangLee last updated on 11/Nov/20

$$\mathrm{Given}\frac{\mathrm{d}{(5t-4)}^2}{\mathrm{d}t}=10\mathrm{g}\left(t\right).$$$$\mathrm{Find}{\int }_1^k\mathrm{g}\left(t\right)\mathrm{d}t\mathrm{in}\mathrm{terms}\mathrm{of}k,\mathrm{where}k$$$$\mathrm{is}\mathrm{a}\mathrm{constant}.$$

Answered by Dwaipayan Shikari last updated on 11/Nov/20

$${\int }_1^{k}du={\int }_1^{k}10g\left(t\right)dtu={(5t-4)}^2$$$${\left[u\right]}_1^k=10{\int }_1^{k}g\left(t\right)dt$$$$\frac{1}{10}({(5k-4)}^2-1)={\int }_1^{k}g\left(t\right)dt=\frac{5}{2}{k}^2-4k+\frac{3}{2}$$$$$$$$$$

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