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Sagot :
It looks like you have
[tex]f(x) = \begin{cases} 3 & \text{if } -4 \le x \le 0 \\ 4 - x^2 & \text{if } 0 < x \le 3\end{cases}[/tex]
and the integral you want to compute is
[tex]\displaystyle \int_{-4}^3 f(x) \, dx[/tex]
Split the integral up at [tex]x=0[/tex]. Then
[tex]\displaystyle \int_{-4}^3 f(x) \, dx = \int_{-4}^0 f(x)\,dx + \int_0^3 f(x)\,dx \\\\ ~~~~~~~~~~~~ = \int_{-4}^0 3 \, dx + \int_0^3 (4 - x^2) \, dx \\\\ ~~~~~~~~~~~~ = 3(0 - (-4)) + \left(4\cdot3 - \frac{3^3}3\right) = \boxed{15}[/tex]
After evaluating the integral we get 15
Integral is the representation of the area of a region under a curve. We approximate the actual value of an integral by drawing rectangles. A definite integral of a function can be represented as the area of the region bounded by its graph of the given function between two points in the line.
Given,
f(x) = 3 if - 4 ≤ x ≤ 0
f(x) = 4 - [tex]x^{2}[/tex] if 0 < x ≤ 3
Then,
[tex]f(x) = \left \{ {{3} \atop {4-x^{2} }}[/tex]
We need to solve the integral -[tex]\int\limits^3_4 {f(x)} \, dx[/tex]
Elaborate the integral with x = 0
-[tex]\int\limits^3_4 {f(x)} \, dx[/tex]
= -[tex]\int\limits^0_4 {f(x)} \, dx + \int\limits^3_0 {f(x)} \, dx[/tex]
= -[tex]\int\limits^0_4 {3} \, dx + \int\limits^3_0 {4-x^{2} } \, dx[/tex]
= 3 ( 0 - (-4)) + (4.3 - [tex]\frac{3^{3} }{3}[/tex])
= 15
Learn more about integrals here :- brainly.com/question/14510286
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