Evaluate the integral `int_0^3(x/2 -1)dx` by interpreting in terms of area.



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oldnick's profile pic

Posted on (Answer #1)

`int_0^3 (x/2-1)dx=[x^2/4-x]_0^3= -3/4`

embizze's profile pic

Posted on (Answer #2)

The line creates two triangles: one from x=0 to 2 and the other from x=2 to 3 bounded by the line and the x-axis.

The area of the triangle from x=0 to 2 : the base is 1 and the height 2 so the area is 1. Since the triangle is beneath the axis we assign it a negative value. Thus its area is -1.

The triangle from x=2 to 3 has base 1/2 and height 1, so its area is 1/4.

The total area is the sum of the triangles' areas: -1+1/4=-3/4.

Using integration:

`int_0^3 (x/2 -1)dx=x^2/4-x|_0^3=9/4-3=-3/4`

The graph:

oldnick's profile pic

Posted on (Answer #3)

The geometric meaning, of integral  is the red stressed area.

oldnick's profile pic

Posted on (Answer #4)

the straigth line  `x/2-1`   has collison with axis `x=0` at the  that is  `y=-1,`  and with the line `x=3`  with `y=1/2`

So the area `A`  is sum of the area of triangle  `T_1`  with base the segment `[0;2]`  and heights the vertical segment `[0;-1]` .

The second triangle is with base the segment `[2;3]`  and heights the vertical segment : `[0;1/2]`   

Then the area `A` is equal:

`A= A_T_1 + A_T_2=1/2 xx (-1) xx 2+ 1/2 xx 1 xx1/2=` `-1+1/4=-3/4`

Just we have got with integral:



rakesh05's profile pic

Posted on (Answer #5)

Given    `int_0^3(x/2-1)dx` .

We know that   the area bounded by the curve `y=f(x)`  and the x-axis is given by

                     Area=`int_a^bf(x)dx` .

So,        `int_0^3(x/2-1)dx` =`[x^2/4-x]`  between 0 to 3



Because area is always positive. So the area is =3/4 square unit.

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