Problem 14 - College Algebra and Calculus: An Applied Approach

Question

Use the values \(\int_{0}^{5} f(x) d x=6\) and \(\int_{0}^{5} g(x) d x=2\) to evaluate the definite integral. (a) \(\int_{0}^{5} 2 g(x) d x\) (b) \(\int_{5}^{0} f(x) d x\) (c) \(\int_{5}^{5} f(x) d x\) (d) \(\int_{0}^{5}[f(x)-f(x)] d x\)

Step-by-Step Solution

Answer

(a) 4, (b) -6, (c) 0, (d) 0

Step 1: Evaluate Integral: \(\int_{0}^{5} 2 g(x) d x\)

Since the integral of \(g(x)\) from 0 to 5 is given as 2, we simply multiply this value by the constant factor 2. \(\int_{0}^{5} 2 g(x) d x = 2 * \int_{0}^{5} g(x) d x = 2 * 2 = 4\)

Step 2: Evaluate Integral: \(\int_{5}^{0} f(x) d x\)

The integral from a to b of a function equals the negative of the integral from b to a of the same function. So, \(\int_{5}^{0} f(x) d x = - \int_{0}^{5} f(x) d x = -6\)

Step 3: Evaluate Integral: \(\int_{5}^{5} f(x) d x\)

The integral from a to a of any function is 0, regardless of the function. So, \(\int_{5}^{5} f(x) d x = 0\)

Step 4: Evaluate Integral: \(\int_{0}^{5}[f(x)-f(x)] d x\)

Inside the integral, we have the difference of \(f(x)\) and \(f(x)\), which is zero. The integral of zero over any interval is 0. So, \(\int_{0}^{5}[f(x)-f(x)] d x = 0\)