3.7: Exercises
Exercise \(\PageIndex{1}\)
Consider the step function \[\Theta(x) = \left\{\begin{array}{ll} 1, &\;\;\;\textrm{for} \; x \ge 0\\ 0,&\;\;\; \textrm{otherwise.}\end{array}\right.\] Write down an expression for the antiderivative of \(\Theta(x)\) , and sketch its graph.
Exercise \(\PageIndex{2}\)
Show that \[\int_0^{2\pi} dx\, [\sin(x)]^2 = \int_0^{2\pi} dx\, [\cos(x)]^2 = \pi.\]
Exercise \(\PageIndex{3}\)
Calculate the following definite integrals:
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\(\displaystyle\int_{0}^\pi dx\; x^2 \sin(2x)\)
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\(\displaystyle\int_{1}^\alpha dx\; x \ln(x)\)
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\(\displaystyle\int_0^\infty dx\;e^{-\gamma x} \, \cos(x)\)
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\(\displaystyle\int_0^\infty dx\;e^{-\gamma x} \, x \cos(x)\)
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\(\displaystyle\int_{-\infty}^\infty dx\;e^{-\gamma |x|}\)
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\(\displaystyle \int_{-\infty}^\infty dx \;e^{-|x+1|} \sin(x)\)
Exercise \(\PageIndex{4}\)
By differentiating under the integral, solve \[\int_0^1 dx\; \frac{x^2-1}{\ln(x)}.\] Hint: replace \(x^2\) in the numerator with \(x^\gamma\) .
- Answer
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Let us define \[I(\gamma) = \int_0^1 \frac{x^\gamma - 1}{\ln(x)},\] so that \(I(2)\) is our desired integral. To take the derivative, first note that \[\frac{d}{d\gamma}(x^\gamma) = \ln(x)\, x^\gamma,\] which can be proven using the generalized definition of the power operation. Thus, \[\begin{align} \frac{d}{d\gamma} I(\gamma) &= \int_0^1 \frac{\ln(x) x^\gamma}{\ln(x)} \\ &= \int_0^1 x^\gamma \\ &= \frac{1}{1+\gamma}.\end{align}\] This can be integrated straightforwardly: \[I(\gamma) = \int \frac{d\gamma}{1+\gamma} = \ln(1+\gamma) + c,\] where \(c\) is a constant of integration, which we now have to determine. Referring to the original definition of \(I(\gamma)\) , observe that \(I(0) = \int_0^1 (1 - 1)/\ln(x) = 0\) . This implies that \(c = 0\) . Therefore, the answer is \[I(2) = \ln(3).\]
Exercise \(\PageIndex{5}\)
Let \(f(x,y)\) be a function that depends on two inputs \(x\) and \(y\) , and define \[I(x) = \int_0^x f(x,y) dy.\] Prove that \[\frac{dI}{dx} = f(x,y) + \int_0^x \frac{\partial f}{\partial x}(x,y) \;dy.\]
Exercise \(\PageIndex{6}\)
Consider the ordinary differential equation \[\frac{dy}{dt} = - \gamma y(t) + f(t),\] where \(\gamma > 0\) and \(f(t)\) is some function of \(t\) . The solution can be written in the form \[y(t) = y(0) + \int_0^t dt' \, e^{-\gamma(t-t')} \, g(t').\] Find the appropriate function \(g\) , in terms of \(f\) and \(y(0)\) .
- Answer
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We are provided with the following ansatz for the solution to the differential equation: \[y(t) = y(0) + \int_0^t dt' e^{-\gamma(t-t')} g(t').\] First, note that when \(t = 0\) , the integral’s range shrinks to zero, so the result is \(y(0)\) , as expected. In order to determine the appropriate function \(g\) , we perform a derivative in \(t\) . The tricky part is that \(t\) appears in two places: in the upper range of the integral, as well as in the integrand. So when we take the derivative, there should be two distinct terms (see problem \(\PageIndex{5}\)): \[\begin{align} \frac{dy}{dt} &= \left[e^{-\gamma(t-t')} g(t')\right]_{t'=t} + \int_0^t dt'(-\gamma) \, e^{-\gamma(t-t')} \, g(t')\\ &= g(t) - \gamma [y(t) - y(0)].\end{align}\] In the last step, we again made use of the ansatz for \(y(t)\) . Finally, comparing this with the original differential equation for \(y(t)\) , we find that \[g(t) - \gamma [y(t) - y(0)] = -\gamma y(t) + f(t) \;\;\; \Rightarrow \;\;\; g(t) = f(t) - \gamma y(0).\] Hence, the solution to the differential equation is \[\begin{align} y(t) &= y(0) + \int_0^t dt' \, e^{-\gamma(t-t')} \,[f(t') - \gamma y(0)] \\ &= y(0)\,e^{-\gamma t} + \int_0^t dt' \, e^{-\gamma(t-t')} f(t').\end{align}\]