If $\int_{1}^{5} f(x) d x=8.1$ and $\int_{3}^{5} f(x) d x=5.7$, find $\int_{1}^{3} f(x) d x$.
$\square$
Need Help?
Read It
Master It

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given the integrals $ \int_{1}^{5} f(x) \, dx = 8.1 $ and $ \int_{3}^{5} f(x) \, dx = 5.7 $, we need to find $ \int_{1}^{3} f(x) \, dx $. ‖ ‖ step 2 ⋮ We can use the property of definite integrals: $ \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 $ \int_{1}^{5} f(x) \, dx = \int_{1}^{3} f(x) \, dx + \int_{3}^{5} f(x) \, dx $. ‖ ‖ step 4 ⋮ Solving for $ \int_{1}^{3} f(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: $ \int_{1}^{3} f(x) \, dx = 8.1 - 5.7 = 2.4 $. ‖ ∻Answer∻ ⚹ $ \int_{1}^{3} f(x) \, dx = 2.4 $ ⚹ ∻Key Concept∻ ⚹Property of Definite Integrals⚹ ∻Explanation∻ ⚹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 $[1, 3]$ by subtracting the integral over $[3, 5]$ from the integral over $[1, 5]$.⚹◊What property of definite integrals was used to solve the problem in the student⍭ Generate me a similar question◊

Sia