# Find the area between f(x) = 3x^2 + 2x and x= 1 and x= 2

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### 4 Answers

f(x) = 3x^2 + 2x and x= 1 and x= 2

First we need to find the difinet integral for f(x) between (1,2)

Let F (x) = intg f(x)

==> F(x) = intg 3x^2 + 2x dx

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

= 3x^3/3 + 2x2 /2 + C

= x^3 + x^2 + C

Now the area is:

A = F(2) - F(1)

= 2^3 + 2^2 + C - (1^3 + 1^2 + C)

= 8 + 4 + C - 1 -1 - c

= 10

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

**A = 10 **

The area under the curve f(x) and between x= a and x = b is given by:

Area = Int f(x) dx from x= a to x = b.

Area = Int {3x^2+2x} dx from x = 1 to x = 2.

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

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

Area = {2^3+2^2 +C}- {1^3+1^2+C}

Area = 12-2 +C-C.

Area = 10.

Therefore the area under the function 3x^2-2x between x =2 and x = 1 is 10 sq units.

Since there is not established the other boundary curve, we'll suppose that we have to calculate the area between f(x), the lines x=1 and x=2 and the x axis.

The definite integral will be calculated with Leibniz-Newton formula:

Int f(x)dx = F(b)-F(a)

We'll calculate the indefinite integral of f(x):

Int f(x)dx = Int (3x^2 + 2x)dx

We'll use the property of integral to be additive:

Int (3x^2 + 2x)dx = Int 3x^2dx + Int 2xdx

Int 3x^2dx = 3*x^3/3 + C

Int 3x^2dx =x^3 + C

Int 2xdx = 2*x^2/2 + C

Int 2xdx = x^2 + C

Int (3x^2 + 2x)dx = x^3 + x^2 + C

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

F(2) - F(1) = 8 + 4 - 2

F(2) - F(1) = 10

**The area bounded by the curve of f(x) and the lines x=1, x=2 and x axis is A=10.**

We have the function. f(x) = 3x^2 + 2x.

To find the required area we need to find the definite integral of 3x^2 + 2x between x = 1 and x =2.

Now the integral of the term x^n is given as x^(n+1) / (n+1)

So Int [3x^2 + 2x]

=> 3*x^3/3 + 2x^2/2

=> x^3 + x^2

Between the limits x=1 and x=2

x^3 + x^2 = 2^3 + 2^2 - 1^3 - 1^2

=> 8 + 4 -1- 1

=> 12 - 2

=> 10

**The required area is 10**