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

Step-by-step solution

Given parabola, y=x^2+3 Tangent of parabola at (3,12) is Y+12 2 =3 x+3 Y=6 x-6 aligned Required area & = ₀^3 (x^2+3 ) d x- ₁^3(6 x-6) d x & = [ x^3 3 +3 x ]₀^3- [3 x^2-6 x ]₁^3 & =(9+9)-((27-18)+3)=18-12=6 aligned