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

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

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:

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

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:

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

`=3^2/4-3`

`=9/4-3=-3/4`

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