# inf f(x) = x^2 + 5x - 3 then find the area between f(x) , x=1 and x=2

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f(x) = x^2 + 5x - 3

Let F(x) = integral f(x)

Then the area between f , x=1 and x= 2 is:

A = F(2) - F(1)

Then , let us determine F(x) first.

F(x) = intg f(x)

= intg (x^2 + 5x -3) dx

= intg x^2 dx + intg 5x dx - intg 3 dx

= x^3/3 + 5x^2 /2 - 3x

F(2) = 8/3 + 20/2 - 6

= ( 16 + 60 - 36)/6 = 40/6 = 20/3

==> F(2) = 20/3

==> F(1) = 1/3 + 5/2 - 3 = (2 + 15 - 18)/6 = -1/6

==> A = F(2) - F(1) = 20/3 - -1/6 = 41/6

**Then the area = 41/6 square units.**

The area A(x) under the curve f(x) and x axis between the ordinates x= 1 and x=2 is given by :

A(x) = Integral f(x) from x= 1 to x=2.

A(x) = Integral {x^2+4x-3}dx from x= 1 to 2.

Area = A(2) -A(1) = Integral(x^2)dx+Integral (5x)dx - Integral 3 dx, from x= 1 to x= 2.

Area = {(1/2+1)x^(2+1) + 5(1/(1+1)x^(1+1) - 3x +C )at x= 2} - {(1/2+1)x^(2+1) + 5(1/(1+1)x^(1+1) - 3x +C )at x= 1}

Area = { 8/3+20/2-6}- {1/3+5/2-3}

Area = 41/6

Or area = 6 5/6

A(x)