The area of the region bounded by the curve y = 3x^2 - 4x + 3, the x-axis, and the vertica… — JAMB UTME

The area of the region bounded by the curve $y = 3x^2 - 4x + 3$ , the $x$ -axis, and the vertical lines $x = 0$ and $x = k$ (where $k > 0$ ) is $18$ square units. What is the value of $k$ ?

Answer: 3

Answer

The value of

k

3

The area under the non-negative curve

y = 3x^2 - 4x + 3

from

x = 0

x = k

is found by calculating

\int_{0}^{k} (3x^2 - 4x + 3) \, dx = k^3 - 2k^2 + 3k

. Setting this expression equal to

18

gives

k^3 - 2k^2 + 3k - 18 = 0

. By the factor theorem, testing

k = 3

yields

3^3 - 2(3)^2 + 3(3) - 18 = 0

. Factoring as

(k - 3)(k^2 + k + 6) = 0

confirms

k = 3

as the only real solution.

Step-by-Step Solution

Set up the definite integral for the area bounded by the curve and the x-axis

\int_{0}^{k} (3x^2 - 4x + 3) \, dx = 18

The curve

y = 3x^2 - 4x + 3

lies entirely above the x-axis for all real

x

because its leading coefficient is positive and its discriminant

(-4)^2 - 4(3)(3) = -20 < 0

Evaluate the definite integral in terms of

k

\left[ x^3 - 2x^2 + 3x \right]_0^k = (k^3 - 2k^2 + 3k) - 0 = k^3 - 2k^2 + 3k

Applying the fundamental theorem of calculus by integrating term by term.

Form and simplify the polynomial equation

k^3 - 2k^2 + 3k - 18 = 0

Equating the definite integral expression to the given area value of 18.

Solve for real values of

k > 0

k = 3

Using the factor theorem on

k^3 - 2k^2 + 3k - 18 = 0

k = 3

yields zero (

27 - 18 + 9 - 18 = 0

). Factoring out

(k - 3)

gives

(k - 3)(k^2 + k + 6) = 0

, where

k^2 + k + 6 = 0

has complex roots.

Key Concept

Determining an unknown boundary limit of a definite integral representing area under a curve

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