Example Question - power function
Here are examples of questions we've helped users solve.
Solving a Definite Integral Involving Exponential and Power Functions
<p>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 \(\int u \, dv = uv - \int v \, du\).</p> <p>Let \( u = x^{1/2} \) and \( dv = e^{2x} \, dx \).</p> <p>Then we have \( du = \frac{1}{2}x^{-1/2} \, dx \) and \( v = \frac{1}{2}e^{2x} \).</p> <p>The integral becomes:</p> <p>\( \int_0^{1/2} x^{1/2}e^{2x} \, dx = \left. \frac{1}{2}x^{1/2}e^{2x} \right|_0^{1/2} - \int_0^{1/2} \frac{1}{2}e^{2x} \frac{1}{2}x^{-1/2} \, dx \)</p> <p>\( = \left. \frac{1}{2}x^{1/2}e^{2x} \right|_0^{1/2} - \frac{1}{4} \int_0^{1/2} x^{-1/2}e^{2x} \, dx \)</p> <p>To solve the remaining integral, we use integration by parts again with \( u = x^{-1/2} \) and \( dv = e^{2x} \, dx \).</p> <p>Then we get \( du = -\frac{1}{2}x^{-3/2}dx \) and \( v = \frac{1}{2}e^{2x} \).</p> <p>The remaining integral is:</p> <p>\( \frac{1}{4} \left( \left. x^{-1/2}e^{2x} \right|_0^{1/2} - \int_0^{1/2} \frac{1}{2}e^{2x}(-\frac{1}{2})x^{-3/2} \, dx \right) \)</p> <p>\(= \frac{1}{4} \left( \left. x^{-1/2}e^{2x} \right|_0^{1/2} + \frac{1}{4} \int_0^{1/2} x^{-3/2}e^{2x} \, dx \right) \)</p> <p>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 \(x\).</p> <p>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.</p>
Calculating the Integral of a Power Function
<p>\[ \int f(x) \, dx = \int x^{-2} \, dx \]</p> <p>\[ = \int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} + C \]</p> <p>\[ = \frac{x^{-1}}{-1} + C \]</p> <p>\[ = -\frac{1}{x} + C \]</p>
Integration of a Power Function
<p>\(\int x^{-2} dx\)</p> <p>\(= \int x^{-2+1}(-2+1)^{-1} dx\)</p> <p>\(= \int x^{-1}(-1)^{-1} dx\)</p> <p>\(= -x^{-1+1}(-1+1)^{-1} + C\)</p> <p>\(= -1 \cdot x^0 + C\)</p> <p>\(= -1 + C\)</p> <p>\(= -1 + C\)</p>
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com