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Magazine Chapter 7: Problem 43
Evaluate the integral by making a \(u\) -substitution and then integrating by parts. $$ \int e^{\sqrt{x}} d x $$
Short Answer
Expert verified
The integral evaluates to \( 2 e^{\sqrt{x}} (\sqrt{x} - 1) + C \).
Step by step solution
01
Choose a substitution
Let's choose a substitution for easier integration. Set \( u = \sqrt{x} \), which means \( x = u^2 \). Therefore, \( dx = 2u \, du \)..
02
Substitute and Simplify
Substitute \( u = \sqrt{x} \) and \( dx = 2u \, du \) into the integral, so the integral becomes: \[ \int e^{u} \, 2u \, du = 2 \int u e^{u} \, du. \]
03
Integrate by Parts
Use integration by parts, choosing \( v = u \) and \( dw = e^u \, du \), hence \( dv = du \) and \( w = e^u \). The integration by parts formula is \( \int v \, dw = vw - \int w \, dv \). Substitute and solve: \[ 2 \int u e^u \, du = 2 \left( u e^u - \int e^u \, du \right) = 2 \left( u e^u - e^u \right). \]
04
Substitute Back
Substitute back \( u = \sqrt{x} \) into the expression: \[ 2 \left( u e^u - e^u \right) = 2 \left( \sqrt{x} e^{\sqrt{x}} - e^{\sqrt{x}} \right) = 2 e^{\sqrt{x}} (\sqrt{x} - 1). \]
05
Finalize the Integral
Don't forget to add the constant of integration \( C \): \[ \int e^{\sqrt{x}} \, dx = 2 e^{\sqrt{x}} (\sqrt{x} - 1) + C. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by Parts is a technique for solving integrals when the product of two functions is involved. It's a powerful tool that is derived from the product rule for differentiation. The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). When applying this method, we strategically choose parts of the integrand as \( u \) and \( dv \), aiming to simplify the resulting integral.
- Choosing \( u \) and \( dv \): Picking the right parts is crucial. A helpful mnemonic is "LIATE" (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), which suggests the priority in selecting \( u \).
- Proceeding with the calculations: Once \( u \) and \( dv \) are chosen, differentiate \( u \) to obtain \( du \), and integrate \( dv \) to find \( v \).
- Substituting and simplifying: Substitute these into the formula and simplify to complete the integration.
U-substitution
U-substitution is a method used to simplify complex integrals, making them easier to evaluate. It involves substituting a part of the integral with a single variable \( u \) in order to transform the original integral into a simpler form. This technique can be thought of as the reverse process of the chain rule for differentiation.
- Selecting \( u \): The choice of \( u \) is crucial. Aim to pick a substitution that simplifies the integral significantly, typically something inside a composite function that complicates the integration.
- Determining \( du \): After choosing \( u \), differentiate it to find \( du \) and rewrite \( dx \) in terms of \( du \).
- Substituting and simplifying: Substitute \( u \) and \( du \) back into the integral, transforming it into a more straightforward form.
Definite and Indefinite Integrals
Integrals are fundamental in calculus, allowing us to find areas, solve equations, and more.
There are two main types: definite and indefinite integrals. Each serves a unique purpose and requires a slightly different approach.
There are two main types: definite and indefinite integrals. Each serves a unique purpose and requires a slightly different approach.
- Indefinite Integrals: Represent the antiderivative of a function and include a constant of integration \( C \). These are used when the limits of integration are not specified.
- Definite Integrals: Calculate the net area under the curve between given limits. They provide a specific numerical value and are independent of the constant \( C \).
