Question

Suppose that \(f\) and \(h\) are continuous functions and that \(\int _ { 1 } ^ { 9 } f ( x ) d x = - 1 , \quad \int _ { 7 } ^ { 9 } f ( x ) d x = 5 , \quad \int _ { 7 } ^ { 9 } h ( x ) d x = 4\) Use the rules in Table 5.3 to find each integral. (a) $$\int _ { 1 } ^ { 9 } - 2 f ( x ) d x \quad \quad$$ (b) $$\int _ { 7 } ^ { 9 } [ f ( x ) + h ( x ) ] d x$$ $$( \mathbf { c } ) \int _ { 7 } ^ { 9 } [ 2 f ( x ) - 3 h ( x ) ] d x \quad$$ (d) $$\int _ { 9 } ^ { 1 } f ( x ) d x$$ (e) $$\int _ { 1 } ^ { 7 } f ( x ) d x \quad$$ (f) $$\int _ { 9 } ^ { 7 } [ h ( x ) - f ( x ) ] d x$$

Short answer

(a) 2, (b) 9, (c) 2, (d) 1, (e) -6, (f) -1

Step-by-step solution

- Solve for (a)

For this part, use the constant multiple rule to pull out the -2 in front of the integral. The solution is \( -2 \times \int _ { 1 } ^ { 9 } f ( x ) d x = -2 \times -1 = 2 \).

- Solve for (b)

For this part, use the sum rule for integrals to separate the integral into the sum of two integrals. The solution is \( \int _ { 7 } ^ { 9 } f ( x ) d x + \int _ { 7 } ^ { 9 } h ( x ) d x = 5 + 4 = 9 \).

- Solve for (c)

For this part, use constant multiple rule and summation rule to distribute the integral. The solution is \( 2 \int _ { 7 } ^ { 9 } f ( x ) d x - 3 \int _ { 7 } ^ { 9 } h ( x ) d x = 2 \times 5 - 3 \times 4 = 2 \).

- Solve for (d)

For this part, use the rule that states swapping the limits of integration negates the integral. This means that the solution is \( -\int _ { 1 } ^ { 9 } f ( x ) d x = -(-1) = 1 \).

- Solve for (e)

We don't have the value of \( \int _ { 1 } ^ { 7 } f ( x ) d x \) directly, but we can get this value by subtracting \( \int _ { 7 } ^ { 9 } f ( x ) d x = 5 \) from \( \int _ { 1 } ^ { 9 } f ( x ) d x = -1 \). So, the solution is \( -1 - 5 = -6 \).

- Solve for (f)

Using the subtraction and reversal rules, we find the solution is \( -\int _ { 7 } ^ { 9 } f ( x ) d x + \int _ { 7 } ^ { 9 } h ( x ) d x = -5 + 4 = -1 \).