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Suppose [tex]$\int_1^3 f(x) \, dx = 5$[/tex], [tex]$\int_3^8 f(x) \, dx = -4$[/tex], and [tex]$\int_3^8 g(x) \, dx = 6$[/tex]. Evaluate the integrals in parts a–d.

a. [tex]$\int_1^3 6 f(x) \, dx = 30$[/tex] (Simplify your answer.)
b. [tex]$\int_3^8 -9 g(x) \, dx = -54$[/tex] (Simplify your answer.)
c. [tex]$\int_3^8 [9 f(x) - g(x)] \, dx = -42$[/tex] (Simplify your answer.)
d. [tex]$\int_8^3 [f(x) + 7 g(x)] \, dx = \square$[/tex] (Simplify your answer.)

Sagot :

Let's solve each part step-by-step using the given values for the integrals [tex]\(\int_1^3 f(x) \, dx = 5\)[/tex], [tex]\(\int_3^8 f(x) \, dx = -4\)[/tex], and [tex]\(\int_3^8 g(x) \, dx = 6\)[/tex].

### Part a: [tex]\(\int_1^3 6f(x) \, dx\)[/tex]
The integral [tex]\(\int_1^3 6f(x) \, dx\)[/tex] can be simplified using the linearity property of integrals:
[tex]\[ \int_1^3 6f(x) \, dx = 6 \int_1^3 f(x) \, dx \][/tex]
Given that [tex]\(\int_1^3 f(x) \, dx = 5\)[/tex]:
[tex]\[ 6 \int_1^3 f(x) \, dx = 6 \times 5 = 30 \][/tex]
Thus, [tex]\(\int_1^3 6f(x) \, dx = 30\)[/tex].

### Part b: [tex]\(\int_3^8 -9g(x) \, dx\)[/tex]
The integral [tex]\(\int_3^8 -9g(x) \, dx\)[/tex] can also be simplified using the linearity property of integrals:
[tex]\[ \int_3^8 -9g(x) \, dx = -9 \int_3^8 g(x) \, dx \][/tex]
Given that [tex]\(\int_3^8 g(x) \, dx = 6\)[/tex]:
[tex]\[ -9 \int_3^8 g(x) \, dx = -9 \times 6 = -54 \][/tex]
Thus, [tex]\(\int_3^8 -9g(x) \, dx = -54\)[/tex].

### Part c: [tex]\(\int_3^8 [9f(x) - g(x)] \, dx\)[/tex]
The integral [tex]\(\int_3^8 [9f(x) - g(x)] \, dx\)[/tex] can be split using the linearity property:
[tex]\[ \int_3^8 [9f(x) - g(x)] \, dx = \int_3^8 9f(x) \, dx - \int_3^8 g(x) \, dx \][/tex]
We can simplify each part individually:
[tex]\[ \int_3^8 9f(x) \, dx = 9 \int_3^8 f(x) \, dx \quad \text{and} \quad \int_3^8 g(x) \, dx \][/tex]
Given that [tex]\(\int_3^8 f(x) \, dx = -4\)[/tex] and [tex]\(\int_3^8 g(x) \, dx = 6\)[/tex]:
[tex]\[ 9 \int_3^8 f(x) \, dx = 9 \times -4 = -36 \][/tex]
Thus,
[tex]\[ \int_3^8 [9f(x) - g(x)] \, dx = -36 - 6 = -42 \][/tex]
Thus, [tex]\(\int_3^8 [9f(x) - g(x)] \, dx = -42\)[/tex].

### Part d: [tex]\(\int_8^3 [f(x) + 7g(x)] \, dx\)[/tex]
The integral [tex]\(\int_8^3 [f(x) + 7g(x)] \, dx\)[/tex] involves reversing the limits of integration. We use the property of integrals that states if we reverse the limits, we need to multiply by [tex]\(-1\)[/tex]:
[tex]\[ \int_8^3 [f(x) + 7g(x)] \, dx = -\int_3^8 [f(x) + 7g(x)] \, dx \][/tex]
We can now split the integral inside:
[tex]\[ \int_3^8 [f(x) + 7g(x)] \, dx = \int_3^8 f(x) \, dx + \int_3^8 7g(x) \, dx \][/tex]
We can further simplify:
[tex]\[ \int_3^8 7g(x) \, dx = 7 \int_3^8 g(x) \, dx \][/tex]
Given that [tex]\(\int_3^8 f(x) \, dx = -4\)[/tex] and [tex]\(\int_3^8 g(x) \, dx = 6\)[/tex]:
[tex]\[ \int_3^8 f(x) \, dx + 7 \int_3^8 g(x) \, dx = -4 + 7 \times 6 = -4 + 42 = 38 \][/tex]
Thus:
[tex]\[ \int_8^3 [f(x) + 7g(x)] \, dx = -\int_3^8 [f(x) + 7g(x)] \, dx = -38 \][/tex]
Thus, [tex]\(\int_8^3 [f(x) + 7g(x)] \, dx = -38\)[/tex].

In conclusion, the evaluated integrals are:
a. [tex]\(\int_1^3 6f(x) \, dx = 30\)[/tex]
b. [tex]\(\int_3^8 -9g(x) \, dx = -54\)[/tex]
c. [tex]\(\int_3^8 [9f(x) - g(x)] \, dx = -42\)[/tex]
d. [tex]\(\int_8^3 [f(x) + 7g(x)] \, dx = -38\)[/tex]
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