Integration is an important concept in calculus where for a given relations and functions, x is a real variable and for the interval, integral will be calculated. Here is one such example.
Topic : Finding an area of the given function by Integration
Definite and indefinite integrals are the two special integrals.
Problem : Given function y = x2 - 5x - 6. Find the area by Integration when function is enclosed by x - axis at the interval (0,4)
Solution :
The area is given by the formula
area = a∫b f(x) dx = a∫b y dx
given y = x2 - 5x - 6
enclosed by x - axis, x = 0 and x = 4
Therefore,
Area = x = 4∫x = 0 y dx
= 0∫4 (x2 - 5x - 6) dx
= [x3/3 - 5x2/2 - 6x x=0]x=4
= 43/3 - 5(4)2/2 - 6(4)
= 64/3 - 80/2 - 24
= (128 - 240 - 144)/6
= (128 - 384)/6
= -256/6
So area is approximately 42.5 squnits
If you have any queries related to the above problem or the concept please feel free to contact calculus help.
Topic : Finding an area of the given function by Integration
Definite and indefinite integrals are the two special integrals.
Problem : Given function y = x2 - 5x - 6. Find the area by Integration when function is enclosed by x - axis at the interval (0,4)
Solution :
The area is given by the formula
area = a∫b f(x) dx = a∫b y dx
given y = x2 - 5x - 6
enclosed by x - axis, x = 0 and x = 4
Therefore,
Area = x = 4∫x = 0 y dx
= 0∫4 (x2 - 5x - 6) dx
= [x3/3 - 5x2/2 - 6x x=0]x=4
= 43/3 - 5(4)2/2 - 6(4)
= 64/3 - 80/2 - 24
= (128 - 240 - 144)/6
= (128 - 384)/6
= -256/6
So area is approximately 42.5 squnits
If you have any queries related to the above problem or the concept please feel free to contact calculus help.
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