# Evaluate definite integral of f(x)=minim{x^2,3x-2} for x=0 to x=2.

neela | High School Teacher | (Level 3) Valedictorian

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To evaluate  int f(x) = min ( x^2, 3x-2)  for x=  0 to x = 2

Solution:

We know  y = x^2 and y = 3x-2 cut at the points  which are  solutions of x^2= 3x-2 or x^2 -3x+2 = 0 . Or (x-2)(x-1) = 0

So y = x^2 and y = 3x-2  intersect at the point whose x coordinates are  at x=1 and x=2.

If you draw the graph , then only  following  will be clear:

So in for 0 < x < 1, min (x^2  , 3x-2)  = 3x -2

And in  1 <x< 2,  , min (x^2 , 3x-2 ) = x^2.

Again in the interval  0 <x <1, y = 3x-2 is below the x axis for 0 <x<2/3

In  the interval 2/3 < x < 1 , 3x-2 is above x axis.

Int f(x) = Int (3x-2) ={ Int (3x-2) dx} from x = 0 to x = 1}+ {Int  x^2 dx from x = 0 to1}

= (3x^2/2-2x  at x= 1 ) -   (3x^2/2-2x  at x= 0 )+( x^3/3 at x  = 2) )+( x^3/3 at x  = 1)

= (3/2-2) + {( 2^3/3 - 1/3)

=  -1/2 +  8/3 -1/3.

= -1/2 +8/3 -1/3 = (-3/6+16/3 -2/6 = 11/6.

Int  min {x^2 , 3x-2} from 0 to 2  is 11/6

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Before evaluating the definite integral, we'll apply the rule of minimum to the function f(x).

minim m(x) = min{u(x),v(x)} is:

- u(x), for u(x)<v(x)

- v(x), for u(x)>v(x)

We'll apply the rule of min to the function f(x):

- x^2, for x^2<3x-2 <=> x^2 - 3x + 2 < 0

- 3x-2, for x^2>3x-2 <=> x^2 - 3x + 2 > 0

We'll calculate the roots of x^2 - 3x + 2 to verify where the expression is positive and where it's negative.

It is obvious that the roots of the expression  x^2 - 3x + 2 are 1 and 2.

x1 + x2 = 3 = S

x1*x2 = 2 = P

1 + 2 = 3

1 * 2 = 2

The xpression is negative over the interval [1,2] and it is positive over the intervals (-inf.,1) U (2,+inf.).

Now, we'll calculate the definite integral.

Int f(x) dx = Int 1 + Int 2

Int 1 = Int f(x)x from x = 0 to x = 1

Int 2 = Int f(x) dx from x = 1 to x = 2

Int 1 = Int (3x-2)dx = Int 3xdx - 2Int dx

Int 1 = 3x^2/2 - 2x

Int 1 = F(1) - F(0)

F(1) - F(0) = 3/2 - 2 = -1/2

Int 1 = -1/2

Int 2 = F(2) - F(1)

Int 2 = Int x^2dx

Int 2 = x^3/3

F(2) - F(1) = 2^3/3 - 1^3/3

F(2) - F(1) = 8/3 - 1/3 = 7/3

Int 2 = 7/3

Int f(x) dx = Int 1 + Int 2

Int f(x) dx = Int 1 + Int 2

Int f(x) dx = -1/2 + 7/3

Int f(x) dx = (-3 + 14)/6

Int f(x) dx = 11/6