Given that \int_{1}^{a} (3x^2 - 2x) \, dx = 48, where a > 1 is a constant, find the value … — JAMB UTME

Given that $\int_{1}^{a} (3x^2 - 2x) \, dx = 48$ , where $a > 1$ is a constant, find the value of $a$ .

Answer: 4

The value of the constant upper limit is 4.

Integrating

3x^2 - 2x

yields

x^3 - x^2

. Applying limits from

1

a

gives

(a^3 - a^2) - (1 - 1) = a^3 - a^2

. Setting

a^3 - a^2 = 48

, solving for

a > 1

gives

a = 4

because

4^3 - 4^2 = 64 - 16 = 48

Integrate the polynomial function with respect to

x

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

Using the power rule of integration

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

for each term.

Evaluate the antiderivative between the upper limit

a

and lower limit

1

[x^3 - x^2]_1^a = (a^3 - a^2) - (1^3 - 1^2) = a^3 - a^2

By the Fundamental Theorem of Calculus,

\int_{b}^{c} f(x)dx = F(c) - F(b)

Equate the expression to the given total integral value and solve for

a

a^3 - a^2 = 48 \implies a = 4

Substituting

a=4

yields

4^3 - 4^2 = 64 - 16 = 48

, which satisfies the equation.

Definite Integrals with Unknown Limits

Question