# Integrals #6 and #7

#### Random Variable

Challenge Problems:

$$\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx$$

$$\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx$$

EDIT: I don't like the second integral. It doesn't even make any sense unless you state that $$\displaystyle \sqrt[3]{2x^{3}-3x^2-x+1}$$ is a real-valued function. Sorry. It's a bad problem.

Moderator edit: Approved Challenge question.

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#### chiph588@

MHF Hall of Honor
Challenge Problems:

$$\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx$$
$$\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx =0$$

A simple Taylor/Laurent expansion of $$\displaystyle e^{x /a}$$ and $$\displaystyle e^{b /x}$$ should prove my assertion.

#### Random Variable

$$\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx =0$$

A simple Taylor/Laurent expansion of $$\displaystyle e^{x /a}$$ and $$\displaystyle e^{b /x}$$ should prove my assertion.
Could you elaborate?

The solution I had in mind is even simpler than that.

#### simplependulum

MHF Hall of Honor
If my calculations are correct , the integrals are both zero .

For the first one ,

Sub. $$\displaystyle x = \frac{ab}{t}$$ in the second integral .

and we will find that it is equal to

$$\displaystyle \int_a^b \frac{e^{x/a}}{x}~dx - \int_a^b \frac{e^{t/a}}{t}~dt$$

And the second integral , i find that it is identical to

$$\displaystyle \sqrt[3]{(2x-1)[(x-\frac{1}{2})^2 - \frac{5}{4}]}$$

Sub. $$\displaystyle x = 1-t$$ and we have

$$\displaystyle I = -I$$

$$\displaystyle I = 0$$

chiph588@

#### Random Variable

If my calculations are correct , the integrals are both zero .

For the first one ,

Sub. $$\displaystyle x = \frac{ab}{t}$$ in the second integral .

and we will find that it is equal to

$$\displaystyle \int_a^b \frac{e^{x/a}}{x}~dx - \int_a^b \frac{e^{t/a}}{t}~dt$$

And the second integral , i find that it is identical to

$$\displaystyle \sqrt[3]{(2x-1)[(x-\frac{1}{2})^2 - \frac{5}{4}]}$$

Sub. $$\displaystyle x = 1-t$$ and we have

$$\displaystyle I = -I$$

$$\displaystyle I = 0$$

Nice.

If for the first integral you made the same substitution but didn't break it up into two integrals, you would get I = -I.

And for the second integral, Maple claims the answer is a complex number and Mathematica doesn't know what to do with it. The issue is the cube root of negative real numbers.

#### chiph588@

MHF Hall of Honor
$$\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx$$
Let $$\displaystyle x=t+\tfrac12$$

We get $$\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx = \int^{\tfrac12}_{-\tfrac12} \sqrt[3]{2(t+\tfrac12)^{3}-3(t+\tfrac12)^2-(t+\tfrac12)+1} \ dx =$$ $$\displaystyle \int^{\tfrac12}_{-\tfrac12} \sqrt[3]{2t^3-\tfrac52t} \ dx = 0$$ since $$\displaystyle \sqrt[3]{2t^3-\tfrac52t}$$ is odd.

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