The area (in sq. units) bounded by the parabo | vedclass

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(A) Given parabola is $y=x^2+3$.First, find the equation of the tangent at $(3,12)$.The derivative is $\frac{dy}{dx} = 2x$. At $x=3$, the slope $m = 2

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$6$

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(A) Given parabola is $y=x^2+3$.First, find the equation of the tangent at $(3,12)$.The derivative is $\frac{dy}{dx} = 2x$. At $x=3$, the slope $m = 2(3) = 6$.The equation of the tangent is $y - 12 = 6(x - 3)$, which simplifies to $y = 6x - 6$.The tangent intersects the $x$-axis at $y=0$, so $6x - 6 = 0 \Rightarrow x = 1$.The region bounded by the parabola, the tangent, and the coordinate axes in the first quadrant is the area under the parabola from $x=0$ to $x=3$ minus the area of the triangle formed by the tangent line, the $x$-axis, and the vertical line $x=3$.Alternatively, the area is $\int_0^3 (x^2+3) dx - \int_1^3 (6x-6) dx$.Calculating the first integral: $\int_0^3 (x^2+3) dx = [\frac{x^3}{3} + 3x]_0^3 = (9 + 9) - 0 = 18$.Calculating the second integral (area of the triangle): $\int_1^3 (6x-6) dx = [3x^2 - 6x]_1^3 = (27 - 18) - (3 - 6) = 9 - (-3) = 12$.Required area = $18 - 12 = 6$ sq. units.

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

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