Let the curve y=y(x) be the solution of the differential equation, dy dx=2(x+1). If the numerical value of area bounded by the curve y=y(x) and…

$$\frac{dy}{dx}=2(x+1)$$dxdy​=2(x+1)

$$\Rightarrow \int dy=\int 2(x+1)dx$$⇒∫dy=∫2(x+1)dx

$$\Rightarrow y(x)={x}^{2}+2x+C$$⇒y(x)=x2+2x+C

Area $=\frac{4\sqrt{8}}{3}$=348​​

$$\Rightarrow 2{\int }_{-1}^{-1+\sqrt{1-C}}(-{(x+1)}^{2}-C+1)dx=\frac{4\sqrt{8}}{3}$$⇒2∫−1−1+1−C​​(−(x+1)2−C+1)dx=348​​

$$\Rightarrow 2{[-\frac{{(x+1)}^{3}}{3}-Cx+x]}_{-1}^{-1+\sqrt{1-C}}=\frac{4\sqrt{8}}{3}$$⇒2[−3(x+1)3​−Cx+x]−1−1+1−C​​=348​​

$$\Rightarrow {(-\sqrt{1-C})}^{3}+3C-3C\sqrt{1-C}-3C-3+3\sqrt{1-C}+3=2\sqrt{8}$$⇒(−1−C​)3+3C−3C1−C​−3C−3+31−C​+3=28​

$$\Rightarrow C=-1$$⇒C=−1

$$\Rightarrow f(x)={x}^{2}+2x-1,f(1)=2$$⇒f(x)=x2+2x−1,f(1)=2