# Evaluating Definite integrals

#### shamieh

##### Active member
Evaluate the following integrals.

a) $\int^1_0 x e^x dx$

So integrating by parts we get

$u = x$ $vu = e^x dx$
$du = dx$ $v = e^x$

$uv - \int vdu = x e^x - \int^1_0 e^x dx$

$$\displaystyle xe^x - e^x |^1_0 = 1$$

b) $$\displaystyle \int^1_0 x^2 e^x \, dx$$

Integrating by parts we get

$$\displaystyle u = x^2$$ $$\displaystyle dv = e^x dx$$
$$\displaystyle du = 2xdx$$ $$\displaystyle v = e^x$$

$$\displaystyle uv - \int vdu = x^2 e^x - \int^1_0 e^x 2x = e^1 - 2$$

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#### Fantini

MHB Math Helper
I would be more careful about notation but your results are correct. Congratulations!

#### shamieh

##### Active member
I would be more careful about notation but your results are correct. Congratulations!
Any idea why the definite integral calculator is somehow getting 2 - e? are they equivalent?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Here is a new approach that you might like

Define

$$\displaystyle f(x) = \int^1_0 e^{xt} \, dt$$

Then integrating with respect to $t$ we have

$$\displaystyle f(x) = \int^1_0 e^{xt} \, dt = \frac{e^{x}-1}{x}$$

Now differentiate both sides with respect ot $x$ we have

$$\displaystyle f'(x) = \int^1_0 t\, e^{xt} \, dt = \frac{xe^{x}-e^x+1}{x^2}$$

Putting $x=1$ we have

$$\displaystyle f'(1) = \int^1_0 t\, e^{t} \, dt = 1$$

Diff w.r.t to $x$ again we have

$$\displaystyle f''(x) = \int^1_0 t^2\, e^{xt} \, dt = \frac{x^2 \,e^{x}-2(xe^x-e^x +1)}{x^3}$$

$$\displaystyle f''(1) = \int^1_0 t^2\, e^{t} \, dt = e-2$$

Any idea why the definite integral calculator is somehow getting 2 - e? are they equivalent?
The answer $2-e<0$ is not possible since the integral must be positive.

#### Prove It

##### Well-known member
MHB Math Helper
No, \displaystyle \begin{align*} 2 - e = - \left( e - 2 \right) \end{align*}. They are not equivalent.

e - 2 is definitely correct. You can tell because the function is always positive, so the area is always above the x axis and thus must be positive.

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#### shamieh

##### Active member
Nevermind I se what I did wrong