# Given the function f(x) = - 2x + 5 calculate the area of the triangle between x,y axis and the graph of f(x)

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f(x) = -2x + 5\

Let F(x) = intg f(x)

x-axis ==> y= 0 ==> x= 5/2

y-axis ==> x= 0

We know that the area is:

A = F(5/2) - F(0)

F(x) = intg f(x)

= intg (-2x + 5) dx

= intg -2x dx + intg 5 dx

= -2x^2/2 + 5x + C

= -x^2 + 5x + C

==> A = F(5/2) - F(0)

= -(5/2)^2 + 5(5/2) - 0

= -25/4 + 25/2 = 25/4

==? The area A = 25/4 square units.

f(x) = -2x+5.

To find the area of the triangle formed with x axis, y axis.

Put x = 0 in f(x) = -2x+5. then f(x) = 5.

Put f(x) = 0 in f(x) = -2x+5. Then 0 = -2x+5. So -2x= -5. Or = x= 5/2.

So f(x) = -2x+5 intercepts x axis at 5/2 , y axis at 5.

So OXY is the right angled triangle , where O is the origin, OX = 5/2 along x axis, and OY = 5 along Y axis.

Therefore the area of the triangle OXY = (1/2)OX*OY = (1/2)(5/2)(5) = 6.25 sq units.

To calculate the area of the triangle ABC, we'll have to determine the x and y intercepts of the graph of the function f(x) = -2x + 5.

To calculate x intercept, we'll have to put y = 0. But y = f(x).

f(x) = 0

-2x + 5 = 0

-2x = -5

x = 5/2

x = 2.5

The intercepting point of the line -2x + 5 and x axis is B(2.5 , 0).

To calculate y intercept, we'll have to put x = 0.

f(0) = -2*0 + 5

f(0) = 5

The intercepting point of the line -2x + 5 and y axis is C(0 , 5).

Since the x axis is perpendicular to y axis, the triangle 0BC is right angled triangle and the area is the half of the product of cathetus OB and OC.

A = OB*OC/2

We'll calculate the lengths of the cathetus OB and OC.

OB = sqrt[(xB-xO)^2 + (yB-yO)^2]

OB = sqrt [(2.5)^2]

OB = 2.5

OC = sqrt[(xC-xO)^2 + (yC-yO)^2]

OC = sqrt 5^2

OC = 5

A = 5*2.5/2

A = 2.5^2

**The area of the right angled triangle OBC is: A = 6.25 square units.**