Question Number 121775 by ZiYangLee last updated on 11/Nov/20
$$\mathrm{Given}\:\frac{\mathrm{d}\left(\mathrm{5}{t}−\mathrm{4}\right)^{\mathrm{2}} }{\mathrm{d}{t}}=\mathrm{10g}\left({t}\right). \\ $$$$\mathrm{Find}\:\int_{\mathrm{1}} ^{{k}} \mathrm{g}\left({t}\right)\mathrm{d}{t}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{k},\:\mathrm{where}\:{k}\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{constant}. \\ $$
Answered by Dwaipayan Shikari last updated on 11/Nov/20
![∫_1 ^k du=∫_1 ^k 10g(t)dt u=(5t−4)^2 [u]_1 ^k =10∫_1 ^k g(t)dt (1/(10))((5k−4)^2 −1)=∫_1 ^k g(t)dt=(5/2)k^2 −4k+(3/2)](Q121777.png)
$$\int_{\mathrm{1}} ^{{k}} {du}=\int_{\mathrm{1}} ^{{k}} \mathrm{10}{g}\left({t}\right){dt}\:\:\:\:\:\:\:\:\:\:\:{u}=\left(\mathrm{5}{t}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\left[{u}\right]_{\mathrm{1}} ^{{k}} =\mathrm{10}\int_{\mathrm{1}} ^{{k}} {g}\left({t}\right){dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{10}}\left(\left(\mathrm{5}{k}−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{1}\right)=\int_{\mathrm{1}} ^{{k}} {g}\left({t}\right){dt}=\frac{\mathrm{5}}{\mathrm{2}}{k}^{\mathrm{2}} −\mathrm{4}{k}+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$