What is the value of the definite integral \int_{1}^{2} \left(3x^2 - \frac{4}{x^2}\right) … — JAMB UTME

What is the value of the definite integral $\int_{1}^{2} \left(3x^2 - \frac{4}{x^2}\right) \, dx$ ?

Integrating

3x^2 - 4x^{-2}

gives

x^3 + \frac{4}{x}

. Substituting the upper bound

x=2

gives

8 + 2 = 10

, and substituting the lower bound

x=1

gives

1 + 4 = 5

. The difference is

10 - 5 = 5

Rewrite the integrand with a negative exponent

3x^2 - 4x^{-2}

Converting reciprocal powers into negative exponents simplifies applying the power rule of integration.

Find the indefinite integral (antiderivative)

\int (3x^2 - 4x^{-2}) \, dx = \frac{3x^3}{3} - \frac{4x^{-1}}{-1} = x^3 + \frac{4}{x}

Apply the power rule

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

to each term.

Evaluate the antiderivative at the upper limit

x = 2

2^3 + \frac{4}{2} = 8 + 2 = 10

Substitute the upper limit into the antiderivative expression.

Evaluate the antiderivative at the lower limit

x = 1

1^3 + \frac{4}{1} = 1 + 4 = 5

Substitute the lower limit into the antiderivative expression.

Subtract the lower limit value from the upper limit value

10 - 5 = 5

According to the Fundamental Theorem of Calculus,

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

Definite Integration of Polynomial and Power Functions

Question