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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 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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