Given the function f(x)=2x+3 if x<0, f(x)=e^x+2, if x>=0, demonstrate that int_0^1 x^2f(x^3)dx=(e+1)/3?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the given definite integral to test if the result coincides with the given result `(e+1)/3` .

You should notice that the limits of integration are larger than 0, hence, you need to select the function `f(x) = e^x + 2` .

Evaluating the function `f(x)` at `x = x^3` yields:

`f(x^3) = e^(x^3) + 2`

Replacing `e^(x^3) + 2` for `f(x^3)` yields:

`int_0^1 x^2 f(x^3)dx = int_0^1 x^2(e^(x^3) + 2) dx `

`int_0^1 x^2(e^(x^3) + 2) dx = int_0^1 (x^2e^(x^3) + 2x^2) dx `

Using the property of linearity of integrals yields:

`int_0^1 x^2(e^(x^3) + 2) dx = int_0^1 (x^2e^(x^3))dx + int_0^1 2x^2 dx`

You need to use the following substitution to evaluate the definite integral `int_0^1 (x^2e^(x^3))dx `

`x^3 = t => 3x^2dx = dt => x^2dx = (dt)/3`

`x = 0= > t = 0`

`x = 1 => t = 1`

`int_0^1 (x^2e^(x^3))dx = int_0^1 e^t *(dt)/3`

`int_0^1 e^t *(dt)/2 = (1/3) e^t|_0^1`

Using the fundamental theorem of calculus yields:

`int_0^1 e^t *(dt)/2 = (1/3) (e^1 - e^0)`

`int_0^1 e^t *(dt)/2 = (1/3)(e - 1)`

You need to evaluate the definite integral `int_0^1 2x^2 dx` , such that:

`int_0^1 2x^2 dx = 2x^3/3|_0^1 => int_0^1 2x^2 dx = 2(1^3/3 - 0^3/3)`

`int_0^1 2x^2 dx = 2/3`

`int_0^1 x^2(e^(x^3) + 2) dx = (1/3)(e - 1) + 2/3`

`int_0^1 x^2(e^(x^3) + 2) dx = (e - 1 + 2)/3`

`int_0^1 x^2(e^(x^3) + 2) dx = (e + 1)/3`

Hence, testing the validity of the statement `int_0^1 x^2(e^(x^3) + 2) dx = (e + 1)/3` yields that `int_0^1 x^2(e^(x^3) + 2) dx = (e + 1)/3` holds.

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