Question - Solving a Definite Integral Involving Exponential and Power Functions
Solution:
To solve the integral \(\int_0^{1/2} x^{1/2}e^{2x} \, dx\), we can use the method of integration by parts, which states that \(\int u \, dv = uv - \int v \, du\).
Let \( u = x^{1/2} \) and \( dv = e^{2x} \, dx \).
Then we have \( du = \frac{1}{2}x^{-1/2} \, dx \) and \( v = \frac{1}{2}e^{2x} \).
The integral becomes:
\( \int_0^{1/2} x^{1/2}e^{2x} \, dx = \left. \frac{1}{2}x^{1/2}e^{2x} \right|_0^{1/2} - \int_0^{1/2} \frac{1}{2}e^{2x} \frac{1}{2}x^{-1/2} \, dx \)
\( = \left. \frac{1}{2}x^{1/2}e^{2x} \right|_0^{1/2} - \frac{1}{4} \int_0^{1/2} x^{-1/2}e^{2x} \, dx \)
To solve the remaining integral, we use integration by parts again with \( u = x^{-1/2} \) and \( dv = e^{2x} \, dx \).
Then we get \( du = -\frac{1}{2}x^{-3/2}dx \) and \( v = \frac{1}{2}e^{2x} \).
The remaining integral is:
\( \frac{1}{4} \left( \left. x^{-1/2}e^{2x} \right|_0^{1/2} - \int_0^{1/2} \frac{1}{2}e^{2x}(-\frac{1}{2})x^{-3/2} \, dx \right) \)
\(= \frac{1}{4} \left( \left. x^{-1/2}e^{2x} \right|_0^{1/2} + \frac{1}{4} \int_0^{1/2} x^{-3/2}e^{2x} \, dx \right) \)
The evaluation of these integrals at the limits \(0\) and \(\frac{1}{2}\) must be approached with caution, because the terms involving \(x^{-1/2}\) and \(x^{-3/2}\) are undefined at \(x=0\). These expressions suggest the integral does not converge at the lower limit, however, this is dependent on the behavior of the exponential function as it approaches zero, which could negate the potential divergence caused by the power of \(x\).
Without performing a limit analysis, the solution remains indeterminate at \(x = 0\). A detailed analysis might involve considering the limit of the integrand as \(x\) approaches zero and applying L'Hôpital's rule if necessary.
Tutor Solution:
The formula for integration by parts is given by:
\[ \int u \, dv = uv - \int v \, du \]First, we need to choose which part of our function will be \( u \) and which will be \( dv \). Let's choose:
\[ u = x^{1/2} \quad \text{and} \quad dv = e^{2x} \, dx \]Next, we find \( du \) and \( v \) by differentiating \( u \) and integrating \( dv \), respectively:
\[ du = \frac{1}{2} x^{-1/2} \, dx \quad \text{and} \quad v = \frac{1}{2} e^{2x} \]We applied the power rule \(\frac{d}{dx} x^n = nx^{n-1}\) to differentiate \(u\), and we used the fact that \(\int e^{ax} \, dx = \frac{1}{a} e^{ax}\) to find \(v\).
Applying the integration by parts formula, we get:
\[ \int_{0}^{1/2} x^{1/2} e^{2x} \, dx = \left. x^{1/2} \cdot \frac{1}{2} e^{2x} \right|_0^{1/2} - \int_{0}^{1/2} \frac{1}{2} e^{2x} \cdot \frac{1}{2} x^{-1/2} \, dx \]This simplifies to:
\[ \frac{1}{2} \left( \left. x^{1/2} e^{2x} \right|_0^{1/2} - \int_{0}^{1/2} x^{-1/2} e^{2x} \, dx \right) \]To evaluate the boundary term, plug in the limits of integration:
\[ \left. x^{1/2} e^{2x} \right|_0^{1/2} = \left(\frac{1}{2}\right)^{1/2} e^{2\cdot(1/2)} - \left(0\right)^{1/2} e^{2\cdot0} \]This evaluates to:
\[ \frac{\sqrt{2}}{2} e^{1} - 0 \]We now have the remaining integral to solve:
\[ \int_{0}^{1/2} x^{-1/2} e^{2x} \, dx \]This integral can be calculated using the same integration by parts method or recognized as a standard integral, depending on one's familiarity with integral tables. However, as our goal was to simplify the original integral, we might end up in a recursive process if the integral does not simplify further. In such cases, considering a different approach or checking for special functions or numerical integration methods might be necessary.
The complexity of this remaining integral suggests that either it could be further simplified or a numerical approach might be needed if we do not find a further simplification. Without additional context or instructions to employ a special function or technique, I will leave the solution at this step.
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