1-80 Evaluate the integral. ∫√(x) e^√(x) d x

Name: Problem 61, Chapter 7: Calculus Early Transcendentals
Uploaded: 2021-12-11T21:07:46-08:00
Duration: 4 min 2 s
Channel: M Z
Description: 1-80 Evaluate the integral. ∫√(x) e^√(x) d x

Key Concepts

Exponential Functions

Exponential functions, characterized by a constant base raised to a variable exponent, have distinctive properties in calculus. They simplify under differentiation and integration, making them central in many integrals. Recognizing an exponential component in an integrand often guides the choice of integration strategies, such as substitution or integration by parts, to effectively handle these functions.

Integration by Parts

Integration by parts is a technique based on the product rule for derivatives and is used to decompose the integral of a product of two functions into simpler parts. It is particularly useful when one function becomes simpler upon differentiation, while the other can be readily integrated. This method systematically reduces the complexity of integrating products, especially when exponential functions are involved.

Substitution Method

The substitution method involves replacing a part of the integrand with a new variable to simplify the integral. By identifying and substituting complex expressions, such as the square root of a variable, the integrand becomes easier to integrate. This technique is pivotal when dealing with composite functions, enabling a clear path to a solution by transforming the differential as well.

Transcript

00:01 Okay, so this is the integral that we would like to look at.

00:04 So we're going to start with just a simple u sub.

00:09 Select u equals square root of x, and then just du is going to be 1 over 2, square root of x, dx, just taking the derivative of both sides.

00:19 And i'm going to bring this over to the other side, and let's observe that square root of x, we already set that equal to u, so we can just say that 2 u, d -u is equal to d -x.

00:31 Okay, so now we want to substitute back in.

00:35 So then we can say that this is equal to two, or the integral of two, and then we've got, we're going to replace this square of x with a u, so we have u squared, and then e to the u, and then just d -u.

00:53 Okay, and i'm just going to quickly bring this two out front, and then now we want to do is we want to do integration by parts.

01:03 So we've already got a u.

01:05 So let let w equal u, or actually a w because u squared, and then dv equal to the rest of that.

01:16 So dv is equal to e to the u, du.

01:19 And then our dw should be 2u, du, and then v, just taking the integral, of this is just going to be e to the u.

01:29 Okay, so now we can apply that that integration by parts, and this should just be equal to two times, well, w times v, so that's u squared, e to the u, minus, and then we need our v time, yeah, the integral of vdw.

01:53 So we've already got this two here...

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