What is the area under #f(x)=5x-1# in #x in[0,2] #?

1 Answer
Mar 9, 2018

Answer:

The net area is #8#, The actual area is #8.2 \ "unit"^2#

Explanation:

We seek the are under #f(x)=5x-1# where #x in [0,2]#

Method 1:

The bounded net area is that of a trapezium with heights:

# f(0) = -1 #
# f(2) = 9 #

and width #2#, So we can use the trapezium formula:

# A=1/2(a+b)h #
# \ \ \ =1/2(-1+9)(2) #
# \ \ \ =8 #

Method 2:

We can use calculus, and evaluate the definite integral:

# A =int_a^b \ f(x) \ dx #
# \ \ \ =int_0^2 \ 5x-1 \ dx #
# \ \ \ =[5/2x^2-x]_0^2 #
# \ \ \ =(20/2-2)-(0-0) #
# \ \ \ =8 #, as before

Note:

Both of the above methods calculate the "net" area, whereas the actual area is somewhat different:

graph{(y-5x+1)(y-10000x)(y-10000x+20000)=0 [-1, 3, -5, 12]}

The actual area is:

# A = 1/2(1/5)(1) + 1/2(9/5)(9) #
# \ \ \ = 1/10 + 81/10 #
# \ \ \ = 82/10 #
# \ \ \ = 8.2 #