Definite integral of $e^\sqrt{x} dx$

Question

I was requested to solve $\int_0 ^1 e^\sqrt{x}\ dx$; I'm fairly new to integrals, and tried both substitution methods and integration by parts. If my attempts would have gotten me far enought, I would show them; the reason I don't is that I wasn't able to go any further with them. Perhaps someone could provide a way to solve this, or at least a hint, anything will be of use. Thank you very much to everybody.

Answer 1

A fairly useful technique when studying integrals that include $x^{1/2}$ to to remember that $$ \frac{d}{dx} x^{1/2} = \frac{1}{2} x^{-1/2} = \frac{1}{2} \frac{1}{x^{1/2}}.$$ Thus the substitution $u = x^{1/2}$ becomes very useful, as the change of variables will once again include the function $x^{1/2}=u$.

Now, looking at your example, we make the change of variable $u(x) = x^{1/2}$, which gives $$ \frac{du(x)}{dx} = \frac{1}{2} \frac{1}{x^{1/2}} = \frac{1}{2u}.$$ Now, being loose with differentials gives $ 2u du = dx,$ and we notice that $u(0) = 0$ and $u(1) = 1$. Therefore, after making the above u-substitution, we obtain $$ \int_0^1 \exp(x^{1/2}) dx = 2\int_0^1 u \exp(u) d u. $$ Now, this last integral can be solved fairly simply through integration by parts, and the fundamental theorem of calculus will finish the computation.

Answer 2

Hint: Do the substitution $x=y^2$ and $\mathrm dx=2y\,\mathrm dy$. You will get$$2\int_0^1ye^y\,\mathrm dy.$$

Answer 3

substitute $$\sqrt(x)=u$$ the integral $$\int ue^udu$$ by partial integration f'=e^u

Answer 4

Hint: Perform a $u$-substitution first, then perform integration by parts.

Let $u=\sqrt{x}$. Then $u^2=x$ and $2u\ du=dx$. \begin{align} \int_0^1 e^{\sqrt{x}}\ dx &= 2\int_0^1 ue^u\ du \\ &=2\bigg(ue^u-\int_0^1e^u\ du\bigg) \\ &= \cdots \textrm{you should be able to get the rest}\end{align}

Answer 5

Looks like everyone else beat me to it. For the sake of still contributing something, I computed an arbitrary definite integral: \begin{align} \int_a^b e^{\sqrt{x}} dx&= x e^{\sqrt{x}} |_a^b - \int_a^b x \frac{e^{\sqrt{x}}}{2\sqrt{x}} dx \\&=xe^{\sqrt{x}}|^b_a - \int_{\sqrt{a}}^{\sqrt{b}} u^2 e^u du \\&=xe^{\sqrt{x}}|^b_a - \left(u^2 e^u |^{\sqrt{b}}_{\sqrt{a}} - 2\int_{\sqrt{a}}^{\sqrt{b}} u e^u du \right) \\&=xe^{\sqrt{x}}|^b_a - \left(u^2 e^u |^{\sqrt{b}}_{\sqrt{a}} - 2\left( u e^u |^{\sqrt{b}}_{\sqrt{a}} - \int_{\sqrt{a}}^{\sqrt{b}} e^u du \right)\right) \\&=xe^{\sqrt{x}}|^b_a - \left(u^2 e^u |^{\sqrt{b}}_{\sqrt{a}} - 2\left( u e^u |^{\sqrt{b}}_{\sqrt{a}} - e^u|^{\sqrt{b}}_{\sqrt{a}} \right)\right) \\&= xe^{\sqrt{x}}|^b_a - x e^{\sqrt{x}} |^{{b}}_{{a}} + 2 \sqrt{x} e^\sqrt{x} |^{{b}}_{{a}} - 2e^{\sqrt{x}}|^{{b}}_{{a}} \\&= 2(\sqrt{x} - 1)e^{\sqrt{x}} |^b_a \end{align} This yields the indefinite integral $$ \int e^{\sqrt{x}} dx = 2(\sqrt{x} - 1)e^{\sqrt{x}} + C $$ which can easily be verified by differentiating.

The moral is that some integrals require some perseverance (most require a great deal more).