- #1

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I would use u=sqroot(x) and du=1/2*sqroot(x), but I'm confused what I would set v=?

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- Thread starter jhayes25
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In summary, the conversation discusses different methods for solving a definite integral of e^sqrt(x) from x=0 to x=1. The first method suggests using u=sqrt(x) and du=1/2*sqrt(x) along with integration by parts. The second method suggests using a variable substitution of u=sqrt(x) and integrating with respect to the new variable. The third method suggests using partial integration. Finally, the conversation suggests using the method of first computing the integral of exp(pt) and then differentiating with respect to the parameter p.

- #1

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- 0

I would use u=sqroot(x) and du=1/2*sqroot(x), but I'm confused what I would set v=?

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- #2

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[tex]\int_0^1 e^{\sqrt x}dx[/tex]

- #3

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rocophysics said:[tex]\int_0^1 e^{\sqrt x}dx[/tex]

yes, haha thanks. I knew I would screw it up if i tried.

- #4

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[tex]t=e^{\sqrt x} \rightarrow \ln t=\sqrt x[/tex] "evaluate for new limits" [tex]|_0^1=|_1^e[/tex]

[tex]dt=\frac{e^{\sqrt x}}{2\sqrt x}dx \rightarrow 2\ln tdt=e^{\sqrt x}dx[/tex]

[tex]2\int_1^e\ln tdt[/tex]

Take it from here.

[tex]dt=\frac{e^{\sqrt x}}{2\sqrt x}dx \rightarrow 2\ln tdt=e^{\sqrt x}dx[/tex]

[tex]2\int_1^e\ln tdt[/tex]

Take it from here.

Last edited:

- #5

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[tex] \int u dv[/tex]

I think you're first going to want to execute the

Remember that in executing substitution you either need to change your limits to those of the new variable (actually the same in this case) or you should rewrite your final integral back in terms of the original variable before you evaluate the limits.

- #6

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Try to do this integral using partial integration. Also try to do it by first computing the integral of exp(p t) where p is an arbitrary parameter and then differentiate both sides w.r.t. p.

- #7

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another way of doing it would be like this

[tex]\int_0^1 e^{\sqrt x}dx=\int_0^1 \sqrt x \frac{e^{\sqrt x}}{\sqrt x }[/tex] And then take the sub

[tex] t=\sqrt x, so,

dt=\frac{dx}{2 \sqrt x}=>2dt=\frac{dx}{\sqrt x}, x=0 => t=0, x=1=> t=1[/tex] then

[tex]\int_0^1 2t e^{t} [/tex] and now you could apply integration by parts, by letting u=t and [tex]v=\int e^{t}dt[/tex]

Integration by parts is a method used in calculus to find the integral of a product of two functions by breaking it down into simpler integrals. It involves using the product rule of differentiation to rewrite the integral as a difference of two terms, and then integrating each term separately.

Integration by parts is typically used when the integral involves a product of two functions, where one function is easy to integrate but the other is not. It is also useful when the integral involves functions that are difficult to integrate using other methods, such as substitution or partial fractions.

The integration by parts formula is ∫u dv = uv - ∫v du, where u and v are the two functions being integrated, and du and dv are their respective differentials. This formula is derived from the product rule of differentiation.

The general rule is to choose the function that is easier to differentiate as the "u" function, and the function that is more difficult to integrate as the "dv" function. However, there are some guidelines and strategies that can be used to make this decision, such as the LIATE rule or the TABULAR method.

Roots can sometimes be involved in the integration by parts process, either as part of the functions being integrated or as a result of integration by parts. For example, when integrating trigonometric functions, the "dv" function may involve a root term that can be integrated using trigonometric identities. Roots may also appear in the solution of the integral as a result of simplifying the "uv" term in the integration by parts formula.

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