# If `int_1^2f(2x+3)dx = 3` , and `int_1^2f(3x+2) dx = 5` ,  then  `int_3^6f(x) dx =`?

lemjay | Certified Educator

`int_3^6 f(x) dx = ?`

To be able to evaluate the above integral, f(x) must be determined.

Let,

`f(x) = ax + c`

where a and c are constants.

Then, replace x with 2x+3 to determine the expression for f(2x+3).

`f(2x+3) = a(2x+3) + c = 2ax + 3a + c`

For f(3x+2), replace the variable x in f(x) with 3x+2.

`f(3x+2) = a(3x+2) + c = 3ax + 2a + c`

Then, substitute the expression of f(2x+3) and f(3x+2) to the given integrals.

`int_1^2 f(2x+3)dx = 3`

` int_1^2 (2ax + 3a + c)dx = 3`

`ax^2 + 3ax + cx|_1^2 = 3`

`[a(2)^2 + 3a(2) + c(2)] - [a(1)^2 + 3a(1) + c(1)] = 3`

`[4a + 6a + 2c] - [a + 3a + c] = 3`

`[10a + 2c] - [4a + c ] = 3`

`6a + c = 3 `      (Let this be EQ. 1.)

For the other integral,

`int_1^2 f(3x + 2) = 5`

`int_1^2 (3ax + 2a + c) dx = 5`

`(3ax^2)/2 + 2ax + cx|_1^2 = 5`

`[(3a(2)^2)/2 + 2a(2) + c(2)] - [(3a(1)^2)/2 + 2a(1) + c(1)] = 5`

`[6a + 4a + 2c] - [(3a)/2 + 2a + c] = 5`

`[10a + 2c] - [(7a)/2 + c] = 5`

`(13a)/2 + c = 5 `     (Let this be EQ.2)

Then, subtract EQ.1 from EQ.2

`(13a)/2 + c = 5`

(-)        `6a + c = 3`

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` 1/2a = 2`

`a = 4`

Substitute the value of a to either EQ.1 or EQ.2. And solve for c.

`6a + c = 3`

` 6(4) + c = 3`

`24 + c = 3`

`c = -21`

Next, substitute the values of and c to f(x).

`f(x) = ax + c = 4x - 21`

Hence, f(x) = 4x - 21 .

Then, evaluate the `int_3^6 f(x)dx` .

`int_3^6 f(x)dx = int_3^6 (4x - 21) dx = 2x^2 - 21x |_3^6`

`= [2(6)^2-21(6)] - [2(3)^2- 21(3)] = [2(36) - 21(6)] - [2(9) - 21(3)]`

`= [72 - 126] - [ 18 - 63] = -54 - (-45) = -9`