Find the value of the definite integral \int_{1}^{2} \left(x + \frac{1}{x^2}\right) dx. — JAMB UTME

Find the value of the definite integral $\int_{1}^{2} \left(x + \frac{1}{x^2}\right) dx$ .

Answer: 2

The value of the definite integral is 2.

Integrating

x + x^{-2}

gives

\frac{x^2}{2} - \frac{1}{x}

. Evaluating from

x=1

x=2

yields

\left(2 - \frac{1}{2}\right) - \left(\frac{1}{2} - 1\right) = \frac{3}{2} - \left(-\frac{1}{2}\right) = 2

Find the antiderivative of

f(x) = x + x^{-2}

F(x) = \frac{x^2}{2} - \frac{1}{x} + C

Apply the power rule of integration

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

for

n \neq -1

Evaluate

F(x)

at the upper boundary

x = 2

F(2) = \frac{2^2}{2} - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}

Substitute

x = 2

into the antiderivative.

Evaluate

F(x)

at the lower boundary

x = 1

F(1) = \frac{1^2}{2} - \frac{1}{1} = \frac{1}{2} - 1 = -\frac{1}{2}

Substitute

x = 1

into the antiderivative.

Calculate

F(2) - F(1)

\frac{3}{2} - \left(-\frac{1}{2}\right) = \frac{3}{2} + \frac{1}{2} = 2

Apply the Fundamental Theorem of Calculus:

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

Definite Integration using the Power Rule

Question