/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Find or evaluate the integral us... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 42

Find or evaluate the integral using substitution first, then using integration by parts. $$ \int_{0}^{2} e^{\sqrt{2 x}} d x $$

Short Answer

Expert verified
The result of the integral \(\int_{0}^{2} e^{\sqrt{2 x}} d x\) is \(2e^2-\frac{1}{2}(e^{2\sqrt{2}}-1)\)

Step by step solution

01

(Substitution)

Start the problem by setting \(u = \sqrt{2x}\). Direct calculation then gives \(du = \frac{dx}{\sqrt{2x}} = \sqrt{2} \frac{dx}{2 \sqrt{x}}\). Cross multiply to isolate \(dx\), getting \(dx = \frac{du}{\sqrt{2}} \sqrt{x}\). Substitute \(u\) back into the expression for \(\sqrt{x}\), giving \(\sqrt{x} = \frac{u}{\sqrt{2}}\). So this gives \(dx = \frac{1}{2} du\). Replace \(dx\) in the original integral and also replace the limits of integration according to \(u\). The new limits will be \(u(0)=0\) and \(u(2)=2\).
02

(Integration by Substitution)

After substitution, the integral becomes: \(\int_0^{2\sqrt{2}} e^u \frac{1}{2} du\). Evaluate the integral as usual getting \(\frac{1}{2}\left[e^u\right]_0^{2\sqrt{2}} = \frac{1}{2}(e^{2\sqrt{2}}-e^0) = \frac{1}{2}(e^{2\sqrt{2}}-1)\).
03

(Integration by Parts)

Now, solve the original integral with the method of parts. First, identify \(u\) and \(dv\) for the integration. Here, we can take \(u = e^{\sqrt{2x}}\) and \(dv = dx\). Now, compute \(du\) and \(v\). Computing \(du\) we get \(du = e^{\sqrt{2x}} \frac{1}{2\sqrt{x}} dx\). And \(v\) will be itself \(x\) as the integral of \(1\) with respect to \(x\) is \(x\).
04

(Apply Integration By Parts Formula)

Applying the formula for integration by parts which is \(\int u dv = uv - \int v du\), our integral becomes: \(\int_{0}^{2} e^{\sqrt{2x}} dx = [xe^{\sqrt{2x}}]_{0}^{2} - \int_{0}^{2} x e^{\sqrt{2x}} \frac{1}{2\sqrt{x}} dx\). Evaluating the first term with the upper and lower limit, we get \(2e^{\sqrt{4}} - 0 = 2e^2\), simplify further to solve the integral.
05

(Solve Remaining Integral Using Substitution)

The remaining integral which needs to be solved is \(- \int_{0}^{2} \frac{x}{2\sqrt{x}} e^{\sqrt{2x}} dx\). Here, again to solve this integral, we can use substitution method form step 1 and 2. After applying the same steps, the integral becomes \(-\frac{1}{2} [e^u]_0^{2\sqrt{2}} = -\frac{1}{2}(e^{2\sqrt{2}}-e^0) =-\frac{1}{2}(e^{2\sqrt{2}}-1)\).
06

(Combine Results)

Add up the result of step 4 and step 5 to get the final result. So, the final result is \(2e^2-\frac{1}{2}(e^{2\sqrt{2}}-1)\)

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