# Find the integral of the function f(x) = (2x^2 -5)/x from x=1 to x= 2.

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### 2 Answers

We have the function f(x) = (2x^2 - 5)/x. We have to find the definite integral from x = 1 to x= 2

f(x) = (2x^2 - 5)/x

=> 2x - 5/x

Int [ f(x) dx] = 2x^2/2 - 5*ln x = x^2 - 5*ln x

In the interval x = 1 to x = 2, we have

2^2 - 5*ln 2 - 1^2 + 5*ln 1

=> 4 - 5*ln 2 - 1 + 0

=> 3 - 5*ln 2

**The required value of the integral of f(x) = (2x^2 - 5)/5 from x=1 to x = 10 = 3 - 5*ln 2**

Given the curve f(x) = (2x^2-5)/x

We need to find the integral of f(x) between 1 and 2.

Let F(x) = Int f(x).

Then the definite integral is :

A = F(2) - F(1)

Let us determine F(x).

==> F(x) = Int f(x) dx = (2x^2-5)/x dx

We will simplify.

==> F(x) = Int (2x^2/x) dx - Int 5/x dx

==> F(x) = Int 2x dx - 5*Int dx/x

==> F(x) = 2x^2/2 - 5*ln x + C

==> F(x) = x^2 - 5lnx + C

==> F(1) = 1- 5ln1 + c = 1

==> F(2) = 4-5ln2 + C

==> A = 4-5ln2 - 1 = 3-5ln2 = -0.47 ( approx.)

**Then the definite integral is 3-5ln2 = -0.47.**