Question

in exercises 7. \\(\int_{-2}^{1} 5 \\, dx\\)

Explanation:

Step1: Recall the integral of a constant

The integral of a constant \( k \) with respect to \( x \) is \( kx + C \) (where \( C \) is the constant of integration). For a definite integral \( \int_{a}^{b} k \, dx \), 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 \( f(x) = 5 \), the antiderivative \( F(x) = 5x \).

Step2: Evaluate the definite integral

Using the Fundamental Theorem of Calculus, we substitute the upper limit \( b = 1 \) and lower limit \( a = -2 \) into \( F(x) \):

\( \int_{-2}^{1} 5 \, dx = F(1) - F(-2) \)

Substitute \( x = 1 \) and \( x = -2 \) into \( F(x) = 5x \):

\( F(1) = 5(1) = 5 \)

\( F(-2) = 5(-2) = -10 \)

Then, \( F(1) - F(-2) = 5 - (-10) = 5 + 10 = 15 \)

Answer:

\( 15 \)