# What is the area under the curve f(x) = (5+x^2)/x and x= 1 and x= 2

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### 3 Answers

To find area under the curve we have to integrate f(x) = (5+x^2)/x between the limits x= 1 and x= 2.

Now Int [ (5+x^2)/x] = Int (5/x) + Int (x^2/x)

=> Int (5/x) + Int (x)

=> 5 ln x + x^2 / 2 + C

For x = 2,

5 ln x + x^2 / 2 + C = 5 ln 2 + 4/2 + C

For x = 1,

5 ln x + x^2 / 2 + C = 5 ln 1 + 1/2 + C

5 ln 2 + 4/2 + C - (5 ln 1 + 1/2 + C)

=> 5 (ln 2 - ln 1) + 2 - 1/2

=> 5 ln 2 + 3/2

**Therefore the area under the curve f(x) = (5+x^2)/x, between the limits x= 1 and x= 2, is 5 ln 2 + 3/2**

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

We need to calculate the area between the curve f(x), x= 1, and x= 2.

We know that the area under the curve f(x) is the integral of f(x).

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

Then the area is:

A = F(2) - F(1).......(1).

Let us determine the integral.

F(x) = intg (5+ x^2 )/x dx

= intg ( 5/x + x) dx

= intg (5/x) dx + intg x dx

= 5*lnx + x^2 /2.

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

**==> F(2) = 5ln2 + 2 + C.**

==> F(1) = 5ln1 + 1/2 + C

But we know that ln1 = 0.

**==> F(1) = 1/2 + C.**

**==> A = F(2) - F(1) = 5ln2 + 2 - 1/2 = 5ln2 + 3/2**

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

**A = 5ln2 + 3/2 square units.**

The area under the given curve and x axis is calculated using Leibniz-Newton formula.

Int (5+x^2)dx/x = F(b) - F(a), where a = 1 and b = 2

First, we'll determine the result of the indefinite integral:

Int (5+x^2)dx/x

We'll use the additive property of integrals:

Int (5+x^2)dx/x = Int 5dx/x + Int x^2dx/x

We'll simplify and we'll draw out the constants and w'ell get:

Int (5+x^2)dx/x = 5Int dx/x + Int xdx

Int (5+x^2)dx/x = 5ln |x| + x^2/2 + C

The resulted expression is F(x).

Now, we'll determine F(b) = F(2):

F(2) = 5ln |2| + 2^2/2

F(2) = 5ln |2| + 2

Now, we'll determine F(a) = F(1):

F(1) = 5ln |1| + 1^2/2

F(1) = 0 + 1/2

We'll determine the area:

A = F(2) - F(1)

A = 5ln |2| + 2 - 1/2

**A = ln 2^5 + 3/2**

**A = ln 32 + 1.5 square units**