The area (in sq. units) bounded by the parabola $y=x^2+3$, the tangent to the parabola at $(3,12)$ and the coordinate axes and lying in the first quadrant is

• Correct answer
  A

  6
• B

  30
• C

  18
• D

  24

Solution

Given parabola, $y=x^2+3$ Tangent of parabola at $(3,12)$ is $\frac{Y+12}{2}=3 x+3 \Rightarrow Y=6 x-6$
$\begin{aligned} \text { Required area } & =\int_0^3\left(x^2+3\right) d x-\int_1^3(6 x-6) d x \\ & =\left[\frac{x^3}{3}+3 x\right]_0^3-\left[3 x^2-6 x\right]_1^3 \\ & =(9+9)-((27-18)+3)=18-12=6\end{aligned}$

Hence, the correct answer is option: 'A'.