Given that \int_{0}^{p} (6x - 4) \, dx = 15 and p > 0, what is the value of p? — JAMB UTME

Given that $\int_{0}^{p} (6x - 4) \, dx = 15$ and $p > 0$ , what is the value of $p$ ?

The value of

p

3

Integrating

6x - 4

yields

3x^2 - 4x

. Substituting the limits from

0

p

gives

3p^2 - 4p

. Equating this to

15

produces the quadratic equation

3p^2 - 4p - 15 = 0

, which factors as

(3p + 5)(p - 3) = 0

. Since

p > 0

, the only valid solution is

3

Find the indefinite integral of the integrand

f(x) = 6x - 4

\int (6x - 4) \, dx = 3x^2 - 4x

Apply the power rule of integration:

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

Evaluate the definite integral from lower limit

0

to upper limit

p

[3x^2 - 4x]_{0}^{p} = (3p^2 - 4p) - (0) = 3p^2 - 4p

Substitute the upper and lower limits into the antiderivative.

Set the evaluated definite integral equal to the given value of

15

and form a quadratic equation.

3p^2 - 4p - 15 = 0

Equate the definite integral value to 15.

Factor the quadratic equation to solve for

p

(3p + 5)(p - 3) = 0 \implies p = -\frac{5}{3}

p = 3

Find two numbers that multiply to

-45

and add up to

-4

, which are

-9

and

5

Select the valid positive value of

p

as specified in the question (

p > 0

p = 3

Disregard the negative root since

p > 0

Determining an unknown boundary limit of a definite integral

Question