# Calculate int_0^1(x+1)^2003dx?

Asked on by ruals

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to evaluate the given definite integral, hence, you need to use integration by substitution, such that:

`x + 1 = t => dx = dt`

You need to change the limits of integration, such that:

`x = 0 => t = 1`

`x = 1 => t = 2`

Replacing the variable and the limits of integration, yields:

`int_1^2 t^2003 dt = (t^2004)/(2004)|_1^2`

You need to use the fundamnetal theorem of calculus, such that:

`int_1^2 t^2003 dt = 1/2004(2^2004 - 1^2004)`

`int_1^2 t^2003 dt = (2^2004 - 1)/2004`

Hence, evaluating the given definite integral, using integration by substitution, yields `int_0^1(x+1)^2003 dx = (2^2004 - 1)/2004` .

Sources:

nick-teal | High School Teacher | (Level 3) Adjunct Educator

Posted on

These other answers are perfectly right, but you do not actually have to do substitution to solve this problem!

The derivative of (x+1) = 1, so there is no chain rule leftovers we need to worry about.  If there was, then we would have to do some sort of substitution.

We can just do a simple power rule integration.

`int_0^1(x+1)^2003dx`  = `[(1/2004)*(x+1)^2004]_0^1`

Then we evaluate.

`(1/2004)*(1 + 1)^2004 - (1/2004)*(0 + 1)^2004`

`(1/2004)*(2)^2004 - (1/2004)*(1)^2004`

`(1/2004)*(2^2004 - 1)`

`(2^2004 - 1)/2004`

The definite integral when evaluated `int_0^1(x+1)^2003dx = (2^2004 - 1)/2004`

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

The definite integral `int_0^1(x+1)^2003dx` has to be determined.

The integral `int x^n dx = x^(n+1)/(n+1)`

In `int_0^1(x+1)^2003dx` the term x+1 is being raised to power 2003.

Let y = x + 1

`dy/dx = 1`

`dy = dx`

Substitute y = x + 1 in the given integral. As y = x +1, the new definite integral is between 1 and 2.

`int_1^2 y^2003 dy`

= `[(1/2004)*y^2004]_1^2`

= `(1/2004)*(2^2004 - 1^2004)`

= `(1/2004)*(2^2004 - 1^2004)`

= `(2^2004 - 1)/2004`

The definite integral ` int_0^1(x+1)^2003dx = (2^2004 - 1)/2004`

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