The area of the region bounded by the parabola y = kx - x^2 (where k > 0) and the x-axis b… — JAMB UTME

The area of the region bounded by the parabola $y = kx - x^2$ (where $k > 0$ ) and the $x$ -axis between its $x$ -intercepts at $x = 0$ and $x = k$ is equal to $36$ square units. What is the value of the positive constant $k$ ?

Cevap: 6

Cevap

The value of the positive constant

k

is 6.

The area bounded by

y = kx - x^2

and the x-axis from

x = 0

x = k

is obtained by integrating

kx - x^2

, which yields

\frac{k^3}{6}

. Setting

\frac{k^3}{6} = 36

gives

k^3 = 216

, whose cube root is

k = 6

Adım Adım Çözüm

Set up the definite integral representing the area bounded by the curve and the x-axis between the intercepts

x = 0

and

x = k

\int_{0}^{k} (kx - x^2) \, dx = 36

The area under a curve

y = f(x)

above the x-axis from

x = a

x = b

is given by

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

Find the antiderivative and evaluate it at the limits

x = k

and

x = 0

\left[ \frac{kx^2}{2} - \frac{x^3}{3} \right]_{0}^{k} = \left(\frac{k(k)^2}{2} - \frac{k^3}{3}\right) - 0 = \frac{k^3}{6}

Applying the integration power rule

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

and simplifying

\frac{k^3}{2} - \frac{k^3}{3} = \frac{k^3}{6}

Set the evaluated expression equal to 36 and solve for

k

\frac{k^3}{6} = 36 \implies k^3 = 216 \implies k = 6

Multiplying both sides by 6 yields

k^3 = 216

, and taking the cube root gives

k = 6

Anahtar Kavram

Definite Integral and Area Under Curve

Soru

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