# How Do You Start Solving the Integral of e^(sqrt(x))?

• BlackMamba
In summary, the conversation discusses solving an integral for the function e^sqrt{x}. The problem is not an improper integral and substitution is suggested to help solve it. One person gets stuck on how to substitute and another suggests using u-substitution and applying integration by parts. The final solution involves new limits of integration at 0 and 16.
BlackMamba
Hello,

I have this integral that is stumping me. Once I get the basic start then I know I can finish the rest myself it's just getting started that's stumping me.

Here is the problem:

$\int_{0}^{4} {e^\sqrt{x}} dx$

First this doesn't appear to me to be an improper integral. So I should just go ahead and solve as normal. But I'm told to find if the integral converges or diverges and evaluate it if it converges. So even though this integral is not an improper integral, I assume I can still apply the same rules.

To begin however, is where I'm getting stuck.

Substitution doesn't appear to help. if I use $u = x^{1/2}$I think I'm making this harder than it needs to be...

Any and all input would be greatly appreciated!

Last edited:
$u^2 = x$, then

$d({u^2})\,=\,2u du$

Thank you

I understand how you got that.

$u=\sqrt{x}$ then,
$u^2 = x$

However, I'm getting lost as to how I'm supposed to substitute that into the original equation.

I think the part that is throwing me is the $d(u^2) = 2udu$

This is as far as I'm getting:

$\int_{0}^{4} {e^\sqrt{u^2}} du$

$\int_{0}^{4} e^u du$

Am I just adding the $2u$ and then do integration by parts?

Last edited:
BlackMamba said:
Thank you

I understand how you got that.

$u=\sqrt{x}$ then,
$u^2 = x$

However, I'm getting lost as to how I'm supposed to substitute that into the original equation.

I think the part that is throwing me is the $d(u^2) = 2udu$

This is as far as I'm getting:

$\int_{0}^{4} {e^\sqrt{u^2}} du$

$\int_{0}^{4} e^u du$

Am I just adding the $2u$ and then do integration by parts?
Using the u-substitution that astronuc showed you, you then have:
$$\int e^{\sqrt{x}}dx = \int e^{\sqrt{u^2}}2u \,du$$

Thanks FrogPad and thanks again to Astronuc.

The help was greatly appreciated!

The new limits of integration will be 0 and 16.

Daniel.

## 1. What is an integral?

An integral is a mathematical concept that represents the area under a curve in a given interval. It is often used to find the total change or accumulation of a quantity over a continuous range of values.

## 2. How do I begin solving an integral?

To begin solving an integral, you need to first determine the limits of integration, which are the starting and ending points of the interval. Then, you can use various integration techniques, such as substitution, integration by parts, or trigonometric identities, to evaluate the integral.

## 3. What are the different types of integrals?

There are two main types of integrals: definite and indefinite. A definite integral has specific limits of integration and gives a numeric value. An indefinite integral does not have limits and represents an entire family of functions.

## 4. What is the purpose of solving an integral?

Solving an integral allows us to find the exact value of a quantity over a continuous range of values. It is used in many fields of science, such as physics, engineering, and economics, to model and analyze real-world situations.

## 5. How can I check if my integral solution is correct?

You can check your integral solution by using differentiation. If you differentiate your solution and get back the original function, then your solution is likely correct. You can also use online tools or graphing calculators to verify your answer.

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