question_answer

The area of the closed figure bounded by \[x=-1,\]\[y=0,\] \[y={{x}^{2}}+x+1\], and the tangent to the curve \[y={{x}^{2}}+x+1\] at A(1,3) is

A) 4/3 sq. units      

B) 7/3 sq. units

C) 7/6 sq. units      

D) None of these

Correct Answer: C

Solution :

[c] Given \[y={{x}^{2}}+x+1={{\left( x+\frac{1}{2} \right)}^{2}}+\frac{3}{4}\] \[\Rightarrow y-\frac{3}{4}={{\left( x+\frac{1}{2} \right)}^{2}}\] This is a parabola with vertex at \[\left( -\frac{1}{2},\frac{3}{4} \right)\]and the curve is concave upwards. \[y={{x}^{2}}+x+1\] or  \[\frac{dy}{dx}=2x+1\]or \[{{\left( \frac{dy}{dx} \right)}_{(1.3)}}=3\] Equation of the tangent at \[A(1,3)\] is \[y=3x\] Required (shaded) area = Area ABDMN \[-\] Area ONA Now, area ABDMN\[=\int\limits_{-1}^{1}{({{x}^{2}}+x+1)dx}\] \[=2\int\limits_{0}^{1}{({{x}^{2}}+1)=\frac{8}{3}}\] Area of ONA\[=\frac{1}{2}\times 1\times 3=\frac{3}{2}.\] \[\therefore \] Required area \[=\frac{8}{3}-\frac{3}{2}\] \[=\frac{16-9}{6}=\frac{7}{6}\]sq. units.