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Math

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PROBLEM

If
abf(x)dx=210f(x)dx+1015f(x)dx25f(x)dx\int_{a}^{b} f(x) d x=\int_{-2}^{10} f(x) d x+\int_{10}^{15} f(x) d x-\int_{-2}^{5} f(x) d x what are the bounds of integration for the first integral?
a=a= and
b=b=

STEP 1

1. The problem involves properties of definite integrals.
2. We need to simplify and combine integrals to find the bounds a a and b b .

STEP 2

1. Simplify the given equation by combining integrals.
2. Identify the bounds a a and b b .

STEP 3

Start by simplifying the right-hand side of the equation. We have:
abf(x)dx=210f(x)dx+1015f(x)dx25f(x)dx\int_{a}^{b} f(x) \, dx = \int_{-2}^{10} f(x) \, dx + \int_{10}^{15} f(x) \, dx - \int_{-2}^{5} f(x) \, dx Combine the integrals on the right-hand side. Notice that:
210f(x)dx25f(x)dx=510f(x)dx\int_{-2}^{10} f(x) \, dx - \int_{-2}^{5} f(x) \, dx = \int_{5}^{10} f(x) \, dx So the equation becomes:
abf(x)dx=510f(x)dx+1015f(x)dx\int_{a}^{b} f(x) \, dx = \int_{5}^{10} f(x) \, dx + \int_{10}^{15} f(x) \, dx

STEP 4

Now, combine the two integrals on the right-hand side:
abf(x)dx=515f(x)dx\int_{a}^{b} f(x) \, dx = \int_{5}^{15} f(x) \, dx

SOLUTION

From the equation abf(x)dx=515f(x)dx\int_{a}^{b} f(x) \, dx = \int_{5}^{15} f(x) \, dx, we can directly identify the bounds:
a=5 a = 5 b=15 b = 15 The bounds of integration for the first integral are:
a=5 a = 5 b=15 b = 15

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