Find the positive constant p such that the value of the definite integral \int_{0}^{p} (3x… — JAMB UTME

Find the positive constant $p$ such that the value of the definite integral $\int_{0}^{p} (3x^2 + 2x) \, dx$ is equal to $12$ .

Answer: 2

The positive constant

p

is equal to

2

Evaluating the definite integral

\int_{0}^{p} (3x^2 + 2x) \, dx

gives

p^3 + p^2

. Setting this expression equal to

12

forms the cubic equation

p^3 + p^2 - 12 = 0

. Solving for the positive real solution yields

p = 2

, since

2^3 + 2^2 = 8 + 4 = 12

Find the antiderivative of the integrand

\int (3x^2 + 2x) \, dx = x^3 + x^2 + C

Applying the power rule for integration

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

term by term gives

\frac{3x^3}{3} + \frac{2x^2}{2} = x^3 + x^2

Evaluate the antiderivative at the limits

0

and

p

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

Using the Fundamental Theorem of Calculus: substitute the upper limit

p

and subtract the value at the lower limit

0

Set up the polynomial equation using the given integral value

p^3 + p^2 = 12 \implies p^3 + p^2 - 12 = 0

The problem states that the definite integral equals 12.

Solve for the positive real value of

p

p = 2

Testing

p=2

2^3 + 2^2 = 8 + 4 = 12

. Factoring gives

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

, where the quadratic factor has a negative discriminant

3^2 - 4(1)(6) = -15 < 0

, making

p=2

the only real solution.

Definite integration with an unknown limit of integration

Question