Question

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. $\frac{11}{6}$
• B. $\frac{7}{6}$
• C. $\frac{5}{6}$
• D. $\frac{1}{6}$

Answer 1

The correct option is C $\frac{5}{6}$

$y=x^2+x+1$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(1,3)}=2\left(1\right)+1=3=m$

Equation of tangent $-:y-3=3(x-1)$

$y=3x$

Area $={\int }_{-1}^0(x^2+x+1)dx+{\int }_0^1(x^2+x+1-3x)dx$

$={[\frac{x^3}{3}+\frac{x^2}{2}+x]}_{-1}^0+{[\frac{x^3}{3}-x^2+x]}_0^1$

$=-(\frac{-1}{3}+\frac{1}{2}+1)+\frac{1}{3}-1+1$

$=\frac{2}{3}-\frac{3}{2}$

$=\frac{-5}{6}$

Hence, the answer is $\frac{5}{6}.$