Question

evaluate the definite integral.\\(\int_{1}^{6} 2 \\, dx = \square\\)

Explanation:

Step1: Recall the integral of a constant

The integral of a constant \( c \) with respect to \( x \) is \( cx + C \) (where \( C \) is the constant of integration). For the definite integral \( \int_{a}^{b} c \, dx \), we use the Fundamental Theorem of Calculus, which states that \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \), where \( F(x) \) is an antiderivative of \( f(x) \).

For \( f(x) = 2 \), the antiderivative \( F(x) = 2x \) (since the derivative of \( 2x \) with respect to \( x \) is \( 2 \)).

Step2: Apply the Fundamental Theorem of Calculus

We evaluate \( F(6) - F(1) \), where \( F(x) = 2x \).

First, find \( F(6) \): \( F(6) = 2 \times 6 = 12 \)

Then, find \( F(1) \): \( F(1) = 2 \times 1 = 2 \)

Now, subtract: \( F(6) - F(1) = 12 - 2 = 10 \)

Answer:

\( 10 \)