# Use elementary geometry to find the area under the graph of y= ½ x + 3 between x= -1 and x= 5. Sketch a picture.

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Now here we have a equation of a straight line. And also we have to lines on the x-axis. We need to find the area which is formed by these lines.

Now let us see the shape of these lines.

y = 1/2x+3

This is a inclined line which is forming a positive angle with x-axis.

x=-1and x=5

These lines are vertical straight lines which goes through points (-1,0) and (5,0) respectively.

Now if we consider the shape of the geometry we obtain from these lines, it is a trapizoid.

We already know the base line and it's points in the trapizoid. They are (-1,0) and (5,0)

Now let's find the points which is on the inclined leg of the trapizoid.

These points are can be obtained by solving y = 1/2x+3 at x=-1 and x=5

y = 1/2x+3 when x=-1 then y = 1/2(-1)+3 = 5/2

y = 1/2x+3 when x=5 then y = 1/2(5)+3 = 11/2

So the points are (-1,5/2),(-1,0),(5,0),(5,11/2)

Let us name these points A,B,C,D respectively.

From these points;

Line AB form by (-1,0) and (-1,5/2) is parallel to line CD form by (5,0) and (5,11/2).

The base or the perpendicular line (BC) is formed by (-1,0) and (5,0).

So area of the trapizoid = (AB+CD)*BC/2

AB = `sqrt((-1-(-1))^2+(5/2-0)^2)` = 5/2

Similarly;

BC = 6

CD = 11/2

So area of the trapizoid = (5/2+11/2)*6/2 = 24

**Area under the graph of y= ½ x+3 between x=-1 and x=5 is 24**.

The following graph shows it's sketch.