A) \[\frac{56}{3}\]
B) \[\frac{32}{3}\]
C) \[\frac{8}{3}\]
D) \[\frac{14}{3}\]
Correct Answer: C
Solution :
\[y={{x}^{2}}-1\] \[y+1)={{x}^{2}}\,\,\,\,\,\,\,\,\,\frac{dy}{dx}\,\,=\,\,2x\] \[{{\left( \frac{dy}{dx} \right)}_{(2,\,\,3)}}\,\,\,\,\,\,\,=\,\,\,\,\,4\] \[(y-3)=4(x-2)\] \[y-\text{ }3=4x\text{ }-\text{ }8\] \[y=\text{ }4x\text{ }-\text{ }5\] \[\int\limits_{-5}^{3}{\left( \frac{y+5}{4} \right)}\,dy\,\,-\,\,\,\,\,\int\limits_{-1}^{3}{\sqrt{y+1\,}dy}\] \[=\,\,\frac{1}{4}\,\left[ \frac{{{y}^{2}}}{2}+5y \right]{{\,}^{3}}_{-5}\,\,-\,{{\left[ 2\frac{{{(y+1)}^{3/2}}}{3} \right]}^{3}}_{-1}\] \[=\,\,\frac{1}{4}\,\left[ \left( \frac{9}{2}+15 \right)-\,\left( \frac{25}{2}\,-\,25 \right) \right]\,\,-\,\frac{2}{3}\,\,[{{(4)}^{3/2}}]\] \[=\,\,\,\,\,\,\frac{1}{4}\,\left[ \frac{39}{2}\,\,+\,\,\frac{25}{2} \right]\,\,-\,\,\frac{2}{3}\,[8]\] \[=\,\,\,\,\,\frac{1}{4}\,\,\times \,\frac{64}{2}\,-\,\,\frac{16}{3}\,\] \[=\,\,\,\,8\,\,-\,\,\frac{16}{3}\,\,=\,\,\frac{8}{3}\]You need to login to perform this action.
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