Find the area of the region bounded by the curve y = 3x^2, the x-axis, and the vertical li… — JAMB UTME

Find the area of the region bounded by the curve $y = 3x^2$ , the $x$ -axis, and the vertical lines $x = 1$ and $x = 3$ .

Answer: 26 square units

The area of the bounded region is 26 square units.

The area under

y = 3x^2

from

x = 1

x = 3

is calculated using the definite integral

\int_{1}^{3} 3x^2 \, dx = [x^3]_{1}^{3} = 3^3 - 1^3 = 27 - 1 = 26

square units.

Set up the definite integral representing the bounded area.

A = \int_{1}^{3} 3x^2 \, dx

The area under a non-negative curve

y = f(x)

from

x = a

x = b

above the

x

-axis is given by the definite integral

\int_{a}^{b} f(x) \, dx

Determine the antiderivative of

3x^2

\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3

Applying the power rule of integration

\int x^n \, dx = \frac{x^{n+1}}{n+1}

gives

x^3

Evaluate the definite integral using the fundamental theorem of calculus.

[x^3]_{1}^{3} = 3^3 - 1^3 = 27 - 1 = 26

Substitute the upper limit

x = 3

and subtract the value of the function evaluated at the lower limit

x = 1

Area under a curve using definite integration

Question