`int_-1^2(1-x)dx` Evaluate the integral by interpreting it in terms of areas.

Textbook Question

Chapter 5, 5.2 - Problem 35 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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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 function in the interval (-1,2). Refer the attached graph.

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

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

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

Area of 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`

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the area enclosed by the curve represented by the function f(x) = (1-x), x axis and the lines x = -1 and x = 2, using the fundamental theorem of calculus, such 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 like terms yields:

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

Hence, evaluating the area enclosed by the curve represented by the function f(x) = (1-x), x axis and the lines x = -1 and x = 2, using the fundamental theorem of calculus, yields `int_(-1)^2 (1 - x) dx = 3/2.`

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