QuestionJuly 28, 2025

If int _(a)^bf(x)dx=int _(-9)^-2f(x)dx+int _(-2)^2f(x)dx-int _(-9)^-5f(x)dx what are the bounds of integration for the first integral? a= square and b= square

If int _(a)^bf(x)dx=int _(-9)^-2f(x)dx+int _(-2)^2f(x)dx-int _(-9)^-5f(x)dx what are the bounds of integration for the first integral? a= square and b= square
If int _(a)^bf(x)dx=int _(-9)^-2f(x)dx+int _(-2)^2f(x)dx-int _(-9)^-5f(x)dx what are the bounds of integration for the first integral? a= square and b= square

Solution
4.1(213 votes)

Answer

a = -5, b = 2 Explanation 1. Analyze the given equation The equation is \int _{a}^{b}f(x)dx=\int _{-9}^{-2}f(x)dx+\int _{-2}^{2}f(x)dx-\int _{-9}^{-5}f(x)dx. We need to find a and b. 2. Simplify the equation Combine the integrals on the right side: \int _{-9}^{-2}f(x)dx + \int _{-2}^{2}f(x)dx - \int _{-9}^{-5}f(x)dx = \int _{-5}^{-2}f(x)dx + \int _{-2}^{2}f(x)dx. 3. Determine bounds a and b The combined integral \int _{-5}^{-2}f(x)dx + \int _{-2}^{2}f(x)dx implies that a = -5 and b = 2.

Explanation

1. Analyze the given equation<br /> The equation is $\int _{a}^{b}f(x)dx=\int _{-9}^{-2}f(x)dx+\int _{-2}^{2}f(x)dx-\int _{-9}^{-5}f(x)dx$. We need to find $a$ and $b$.<br /><br />2. Simplify the equation<br /> Combine the integrals on the right side: $\int _{-9}^{-2}f(x)dx + \int _{-2}^{2}f(x)dx - \int _{-9}^{-5}f(x)dx = \int _{-5}^{-2}f(x)dx + \int _{-2}^{2}f(x)dx$.<br /><br />3. Determine bounds $a$ and $b$<br /> The combined integral $\int _{-5}^{-2}f(x)dx + \int _{-2}^{2}f(x)dx$ implies that $a = -5$ and $b = 2$.
Click to rate:

Similar Questions