Calculate the area of the finite region bounded by the parabola y = 3x^2 - 12x + 9 and the… — JAMB UTME

Calculate the area of the finite region bounded by the parabola $y = 3x^2 - 12x + 9$ and the $x$ -axis.

Answer: 4 square units

The area of the bounded region is 4 square units.

Finding the x-intercepts of

y = 3x^2 - 12x + 9

gives

x = 1

and

x = 3

. Integrating

-y

from

1

3

gives

\left[-x^3 + 6x^2 - 9x\right]_{1}^{3} = 0 - (-4) = 4

square units.

Determine the limits of integration by finding the x-intercepts of the curve.

x = 1

and

x = 3

The bounded region lies between the points where the curve intersects the x-axis (

y = 0

Set up the definite integral with the correct integrand sign.

A = \int_{1}^{3} (9 - 12x + 3x^2) \, dx = \int_{1}^{3} (-3x^2 + 12x - 9) \, dx

Since

y \le 0

[1, 3]

, negating the function ensures the calculated area is positive.

Integrate term-by-term and evaluate between upper limit 3 and lower limit 1.

\left[-x^3 + 6x^2 - 9x\right]_{1}^{3} = (0) - (-4) = 4

Applying the Fundamental Theorem of Calculus yields the exact value of 4.

Area bounded by a curve and the x-axis lying below the x-axis

Question