Q. 26

# Using integration

Answer :

Given,

• ABC is a triangle

• Equation of side AB of y = 2x + 1

• Equation of side BC of y = 3x+1

• Equation of side CA of x=4

By solving AB & BC we get the point B,

AB : y = 2x + 1 , BC: y = 3x + 1

2x + 1 = 3x + 1

x= 0

by substituting x = 0 in AB we get y = 1

The point B = (0,1)

By solving BC & CA we get the point C,

AC : x = 4 , BC: y = 3x + 1

y = 12 + 1

y = 13

The point C = (4,13)

By solving AB & AC we get the point A,

AB : y = 2x+1 , AC : x = 4

y = 8 +1

y = 9

The point A = (4,9)

These points are used for obtaining the upper and lower bounds of the integral.

From the given information, the area under the triangle (colored) can be given by the below figure. From above figure we can clearly say that, the area between ABC is the area to be found.

The required area is

Area of ABC = Area under OBCD – Area under OBAD

Now, the combined area under the rABC is given by

Area under rABC

=Area under AB + Area under BC - Area under AC

Area of the rABC=    = 24+4 – 20 = 8

Therefore, area of the rABC is 8 sq.units.

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