Question

Find the area bounded by the curve $y=x^2+x+1$ and tangent to it at $(1,3)$ from $x=-1$ to $x=1$

• A. $\frac{2}{3}$
• B. $\frac{5}{3}$
• C. $-\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{2}{3}$

$y=x^2+x+1$

Equation of tangent at $(1,3)$

$3=x+1+1$

$x=1$

Area between curve & tangent from $x=-1$ to $1$

$A={\int }_{-1}^1(x^2+x+1-1)dx$

$A={\int }_{-1}^1(x^2+x)dx$

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

$A=\frac{1}{3}+\frac{1}{2}+\frac{1}{3}-\frac{1}{2}=\frac{2}{3}$