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You are watching: Evaluate the integral by interpreting it in terms of areas.



`int_-1^2(1-x)dx`

To interpret the integral in terms of area , graph the integrand.

The integrand is the function `f(x)=1-x`

Graph the role in the term (-1,2). To express the fastened graph.

The bounded region forms 2 triangles, one triangle below the x-axis and 2nd triangle above the x-axis.

Area of triangle above...


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`int_-1^2(1-x)dx`

To interpret the integral in terms of area , graph the integrand.

The integrand is the function `f(x)=1-x`

Graph the role in the interval (-1,2). Express the enclosed graph.

The bounded region forms 2 triangles, one triangle below the x-axis and second triangle over the x-axis.

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Area of triangle above the x-axis `A_2=1/2b_2h_2`

`A_2=1/2*2*2=2`

Area that triangle below the x-axis `A_1=1/2b_1h_1`

`A_1=1/2*1*1=1/2`

So,`int_-1^2(1-x)dx=A_2-A_1`

`=2-1/2`

`=3/2`

 


*


You need to evaluate the area enclosed by the curve stood for by the function f(x) = (1-x), x axis and also the present x = -1 and x = 2, making use of the fundamental theorem that calculus, together that:

`int_(-1)^2 (1 - x) dx = int_(-1)^2 dx - int_(-1)^2 x dx`

`int_(-1)^2 (1 - x) dx = (x - x^2/2)|_(-1)^2`

`int_(-1)^2 (1 - x) dx = (2 - 2^2/2) - (-1) + (-1)^2/2) `

`int_(-1)^2 (1 - x) dx = (2 - 2 + 1 + 1/2)`

Reducing choose terms yields:

`int_(-1)^2 (1 - x) dx = 3/2`

Hence, assessing the area enclosed by the curve stood for by the duty f(x) = (1-x), x axis and also the lines x = -1 and also x = 2, utilizing the an essential theorem of calculus, returns `int_(-1)^2 (1 - x) dx = 3/2.`


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