# Evaluate the following Integrals over the given intervals: A) (4/2x-1) - (3/x+4) over [1,3] B) cosh 3x-sinh4x over [0,Ln2]

sciencesolve | Teacher | (Level 3) Educator Emeritus

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a) You need to evaluate definite integral `int_1^3 (4/(2x-1) - 3/(x+4))dx,`  hence you should split the integral into two simpler integrals such that:

`int_1^3 (4/(2x-1) - 3/(x+4))dx = int_1^3 (4/(2x-1)dx - int_1^3 3/(x+4))dx`

You need to solve the definite integral `int_1^3 4/(2x-1)dx ` , hence you should come up with the substitution `2x - 1 = y`  such that:

`2x - 1 = y =gt 2dx = dy =gt dx = (dy)/2`

`int 4/(2x-1)dx = 4 int (dx)/(2x-1)`

`4 int (dx)/(2x-1) = 4/2 int (dy)/(y) = 2ln |y| + c`

`int_1^3 4/(2x-1)dx = 2ln (2x-1)|_1^3`

`int_1^3 4/(2x-1)dx = 2(ln(6-1) - ln(2-1))`

`int_1^3 4/(2x-1)dx = 2(ln(5) - ln(1))`

`int_1^3 4/(2x-1)dx = ln 5^2 = ln 25`

`int_1^3 3/(x+4)dx = 3ln(x+4)|_1^3`

`int_1^3 3/(x+4)dx = 3(ln(3+4) - ln(1+4))`

`int_1^3 3/(x+4)dx = 3(ln(7/5))`

`int_1^3 3/(x+4)dx = ln (7/5)^3`

`int_1^3 (4/(2x-1) - 3/(x+4))dx = ln 25 - ln (7/5)^3`

`int_1^3 (4/(2x-1) - 3/(x+4))dx = ln 3125/343`

Hence, evaluating definite integral yields `int_1^3 (4/(2x-1) - 3/(x+4))dx = ln 3125/343.`

b) You need to evaluate definite integral `int_0^ln2 (cosh3x-sinh4x) dx` , hence you should split the integral into simpler integrals such that:

`int_0^ln2 (cosh3x-sinh4x)dx = int_0^ln2 (cosh3x) dx -int_0^ln2 (sinh4x) dx`

`int_0^ln2 (cosh3x-sinh4x) dx = (1/3*(sinh 3x) - 1/4 (cosh 4x))|_0^ln 2`

`int_0^ln2 (cosh3x-sinh4x) dx = 1/3(e^(3x) - e^(-3x))/2|_0^ln 2 - 1/4(e^(4x) + e^(-4x))/2|_0^ln2`

`int_0^ln2 (cosh3x-sinh4x) dx = 1/3(e^(ln8) - e^(ln(1/8)))/2 - 1/4(e^(ln16) + e^(ln(1/16)) - 1 - 1)/2`

`int_0^ln2 (cosh3x-sinh4x) dx = (1/6)(8 - 1/8) - (1/8)(16 + 1/16 - 2)`

`int_0^ln2 (cosh3x-sinh4x) dx = 63/48 - 225/128`

`int_0^ln2 (cosh3x-sinh4x) dx = (504 - 675)/384`

`int_0^ln2 (cosh3x-sinh4x) dx = -171/384`

Hence, evaluating definite integral over given interval yields `int_0^ln2 (cosh3x-sinh4x) dx = -171/384.`