What is the area of the region bounded by the curve y = 6x - x^2 and the line y = 2x? — JAMB UTME

What is the area of the region bounded by the curve $y = 6x - x^2$ and the line $y = 2x$ ?

\frac{32}{3}

square units

Equating

y = 6x - x^2

and

y = 2x

gives intersection points

x = 0

and

x = 4

. Integrating the upper curve minus the lower line,

\int_{0}^{4} (4x - x^2) \, dx = \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4} = 32 - \frac{64}{3} = \frac{32}{3}

square units.

Find the points of intersection between the curve and the line

x = 0

and

x = 4

Set

6x - x^2 = 2x \implies 4x - x^2 = 0 \implies x(4 - x) = 0

to find the integration bounds.

Set up the definite integral for the area between the two curves

A = \int_{0}^{4} ((6x - x^2) - 2x) \, dx = \int_{0}^{4} (4x - x^2) \, dx

The area between two functions

f(x)

and

g(x)

from

x=a

x=b

is given by

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

where

f(x) \ge g(x)

[a, b]

Compute the indefinite integral of the integrand

\int (4x - x^2) \, dx = 2x^2 - \frac{x^3}{3}

Apply the standard power rule of integration

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

Evaluate the definite integral using the limits

0

and

4

\left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4} = \left( 2(4)^2 - \frac{4^3}{3} \right) - 0 = 32 - \frac{64}{3} = \frac{32}{3}

square units

Substitute the upper limit

x = 4

and lower limit

x = 0

and simplify fractions.

Area Between Two Curves

Estimated Time:2m 0s

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