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