What is the area of the region bounded by the curve y = x^2 - 4, the x-axis, and the lines… — JAMB UTME

What is the area of the region bounded by the curve $y = x^2 - 4$ , the $x$ -axis, and the lines $x = 0$ and $x = 3$ ?

\frac{23}{3}

square units

The curve

y = x^2 - 4

intersects the x-axis at

x = 2

. To find the total enclosed area between

x = 0

and

x = 3

, the integral must be split into two parts:

[0, 2]

, where the curve is below the x-axis (yielding an area of

\frac{16}{3}

), and

[2, 3]

, where the curve is above the x-axis (yielding an area of

\frac{7}{3}

). Summing these absolute values gives

\frac{16}{3} + \frac{7}{3} = \frac{23}{3}

square units.

Find the x-intercept of the curve

y = x^2 - 4

within the interval

[0, 3]

x^2 - 4 = 0 \implies x = 2

. The curve lies below the x-axis for

0 \le x < 2

and above the x-axis for

2 < x \le 3

Total geometric area requires evaluating regions below and above the x-axis separately so negative integral values do not cancel positive area.

Calculate the area

A_1

of the region below the x-axis from

x = 0

x = 2

A_1 = \left| \int_{0}^{2} (x^2 - 4) \, dx \right| = \left| \left[ \frac{x^3}{3} - 4x \right]_{0}^{2} \right| = \left| \frac{8}{3} - 8 \right| = \left| -\frac{16}{3} \right| = \frac{16}{3}

square units.

The curve is below the x-axis, so taking the absolute value gives the true physical area.

Calculate the area

A_2

of the region above the x-axis from

x = 2

x = 3

A_2 = \int_{2}^{3} (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_{2}^{3} = (9 - 12) - \left( \frac{8}{3} - 8 \right) = -3 - \left(-\frac{16}{3}\right) = \frac{7}{3}

square units.

The curve lies above the x-axis on this interval, yielding a positive definite integral.

Sum the areas of the two regions to find the total bounded area.

\text{Total Area} = A_1 + A_2 = \frac{16}{3} + \frac{7}{3} = \frac{23}{3}

square units.

Adding the individual positive areas yields the total bounded area.

Area bounded by a curve that crosses the x-axis

Question