Respuesta :
Answer:
\int \limits_{-2}^{4} f(x) \, dx
Step-by-step explanation:
The objective is to write
[tex]\int \limits_{-3}^{1} f(x) \, dx + \int \limits_{1}^{4} f(x) \, dx - \int \limits_{-3}^{-2} f(x) \, dx[/tex]
as a single integral in the form
[tex]\int \limits_{a}^{b} f(x)\, dx[/tex].
We consider the segments [tex][-3, 1], [1, 4][/tex] and [tex][-3,-2][/tex]. If we combine the first and the second segment, we obtain
[tex][-3,1] \cup [1,4] = [-3,4][/tex]
Therefore, adding the first two integrals gives
[tex]\int \limits_{-3}^{1} f(x) \, dx + \int \limits_{1}^{4} f(x) \, dx = \int \limits_{-3}^{4} f(x) \, dx[/tex]
Now,we have
[tex]\int \limits_{-3}^{4} f(x) \, dx - \int \limits_{-3}^{-2} f(x) \, dx[/tex]
To subtract them, we need to find the difference of the segments [tex][-3,4][/tex] and [tex][-3,-2][/tex].
[tex][-3,4] \; \backslash \; [-3,-2] = [-2,4][/tex]
Therefore,
[tex]\int \limits_{-3}^{4} f(x) \, dx - \int \limits_{-3}^{-2} f(x) \, dx = \int \limits_{-2}^{4} f(x) \, dx[/tex]
Thus, the single integration of all the definite integral is mentioned below:
[tex]\int_{-2}^{4} f(x) dx[/tex]
Given the integral is,
[tex]\int_{-3}^{1} f(x) dx + \int_{1}^{4} f(x) dx - \int_{-3}^{-2} f(x)dx[/tex]
We need to change the whole definite integral into a single integral.
Now, we have the limits of integrations are [ - 3, 1 ], [ 1, 4 ], and [ - 3, -2 ].
Now, the first limit and second limit of integration are in addition, therefore combining forms of the integration are the union of these units.
Thus,
[ - 3, 1 ] U [ 1, 4 ] = [ -3 , 4 ]
Now, the second limit and third limit of integration are in subtraction, therefore combining forms of the integration are the intersection of these units.
Thus,
[ 1, 4 ] intersection [ - 3, -2 ] = [ null set]
Now, the combination of [ -3 , 4 ] and [ -null set] is [ - 2, 4].
Thus, the single integration of all the definite integral is mentioned below:
[tex]\int_{-2}^{4} f(x) dx[/tex]
To know more about the integrals, please refer to the link:
https://brainly.com/question/22008756
