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Posted about 1 month ago

If 15f(x)dx=8.1\int_{1}^{5} f(x) d x=8.1 and 35f(x)dx=5.7\int_{3}^{5} f(x) d x=5.7, find 13f(x)dx\int_{1}^{3} f(x) d x.
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Answer from Sia
Posted about 1 month ago
Solution by Steps
step 1
Given the integrals 15f(x)dx=8.1 \int_{1}^{5} f(x) \, dx = 8.1 and 35f(x)dx=5.7 \int_{3}^{5} f(x) \, dx = 5.7 , we need to find 13f(x)dx \int_{1}^{3} f(x) \, dx
step 2
We can use the property of definite integrals: abf(x)dx+bcf(x)dx=acf(x)dx \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx
step 3
Applying this property, we have 15f(x)dx=13f(x)dx+35f(x)dx \int_{1}^{5} f(x) \, dx = \int_{1}^{3} f(x) \, dx + \int_{3}^{5} f(x) \, dx
step 4
Solving for 13f(x)dx \int_{1}^{3} f(x) \, dx : 13f(x)dx=15f(x)dx35f(x)dx \int_{1}^{3} f(x) \, dx = \int_{1}^{5} f(x) \, dx - \int_{3}^{5} f(x) \, dx
step 5
Substituting the given values: 13f(x)dx=8.15.7=2.4 \int_{1}^{3} f(x) \, dx = 8.1 - 5.7 = 2.4
13f(x)dx=2.4 \int_{1}^{3} f(x) \, dx = 2.4
Key Concept
Property of Definite Integrals
The property of definite integrals allows us to break down or combine integrals over adjacent intervals. In this case, we used it to find the integral over [1,3][1, 3] by subtracting the integral over [3,5][3, 5] from the integral over [1,5][1, 5].

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