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. $\frac{4}{3}\text{sq. units}$
• B. $\frac{7}{3}\text{sq. units}$
• C. $\frac{7}{6}\text{sq. units}$
• D. None of these

Answer 1

The correct option is C $\frac{7}{6}\text{sq. units}$

Given
$x^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}$
$\Rightarrow y-\frac{3}{4}={(x+\frac{1}{2})}^2$
This is a parabola with vertex at $(-\frac{1}{2},\frac{3}{4})$ and the curve is concave upwards.

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

Equation of the tangent at $A(1,3)$ is $y=3x$


Required (shaded) area, $S=\text{Area ABDMN}-\text{Area of}△ONA$

Now,
$\text{Area ABDMN}=\underset{-1}{\overset{1}{\int }}(x^2+x+1)dx$
$=2{\int }_0^1(x^2+1)(x\to \text{odd function})$
$=\frac{8}{3}$

$\text{Area of}△ONA=\frac{1}{2}\times 1\times 3=\frac{3}{2}$

$\therefore S=\frac{8}{3}-\frac{3}{2}=\frac{7}{6}\text{sq. units}$