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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.1.18
Chapter 8, Problem 8.1.18

The integrals in Exercises 1–44 are in no particular order. Evaluate each integral using any algebraic method, trigonometric identity, or substitution you think is appropriate.
∫ (2^(√y) dy) / 2√y

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1
Start by examining the integral: \(\int \frac{2^{\sqrt{y}}}{2\sqrt{y}} \, dy\). Notice that the expression involves \(\sqrt{y}\) both in the exponent and the denominator, suggesting a substitution involving \(\sqrt{y}\) might simplify the integral.
Let \(u = \sqrt{y} = y^{1/2}\). Then, differentiate both sides with respect to \(y\) to find \(du\) in terms of \(dy\): \(u = y^{1/2} \implies du = \frac{1}{2\sqrt{y}} dy\).
Rearrange the differential to express \(dy\) in terms of \(du\): from \(du = \frac{1}{2\sqrt{y}} dy\), multiply both sides by \(2\sqrt{y}\) to get \(dy = 2\sqrt{y} \, du\).
Substitute \(u\) and \(dy\) back into the integral: replace \(2^{\sqrt{y}}\) with \$2^u\( and \)dy$ with \(2\sqrt{y} \, du\). Notice that the \(2\sqrt{y}\) in the denominator and numerator will cancel out, simplifying the integral to \(\int 2^u \, du\).
Now, integrate \(\int 2^u \, du\) using the formula for integrating exponential functions with base \(a\): \(\int a^u \, du = \frac{a^u}{\ln(a)} + C\). Apply this formula with \(a=2\) to find the antiderivative in terms of \(u\), then substitute back \(u = \sqrt{y}\) to express the answer in terms of \(y\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables. It involves identifying a part of the integrand as a new variable, differentiating it, and rewriting the integral in terms of this variable to make it easier to evaluate.
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Substitution With an Extra Variable

Exponential Functions with Variable Exponents

Exponential functions where the exponent is a function of the variable, such as 2^(√y), require careful handling. Understanding how to differentiate and integrate such functions often involves rewriting the expression using properties of exponents and logarithms.
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Exponential Functions

Algebraic Manipulation of Integrands

Algebraic manipulation involves rewriting the integrand to a more convenient form before integrating. This can include factoring, simplifying radicals, or expressing terms in a way that aligns with substitution or known integral formulas.
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Completing the Square to Rewrite the Integrand
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