Prove:
$$ \int_{0}^{1} e^{x} \sqrt{2 x+1}\left(1+x+x^{2}\right)^{100} d x \leq \sqrt{\frac{\left(e^{2}-1\right)\left(3^{201}-1\right)}{402}} $$
You may not use any integral computation tools such as Wolfram Alpha for this question. HINT: Think about inner products.
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Solve the integral substitute the value of picture below please i need some help i'm desperate please pleaseI hace a Lot homework
\(\frac{\mathscr{L}}{\frac{\rho_{\infty} V_{\infty}^{2} c}{2}}=\int_{0}^{\frac{b}{2}}\left(\frac{\partial C_{l}}{\partial \alpha}\left(\alpha-\alpha_{C_{l}=0}\right)+\frac{\partial C_{l}}{\partial \alpha} \varphi(v)-\frac{\partial C_{l}}{\partial \alpha} \frac{\dot{p} v}{V_{\infty}}+\frac{\partial C_{l}}{\partial \beta} \beta \mathscr{U}_{b_{2}}(v)\right) v d v\)substitute this value in the integral and solve:
1-Given the function:
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math
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Round your answers to two decimal places.
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Let \(\left\{\varphi_{n}(x)\right\}\) be an orthogonal set of functions on \([a, b]\) such that \(\psi_{0}(x)=1\) and \(\varphi_{1}(x)=x\), Show that \(\int_{a}^{b}(\alpha x+\beta) \varphi_{n}(x) d x=0\) for \(n=2,3, \ldots\) and any constants \(\alpha\) and \(\beta\),First we note that \(\alpha x+\beta=(\square) \Phi_{1}(x)+(\square \quad) \Psi_{0}(x)\).Using this together with the fact that \(\varphi_{0}\) and \(\varphi_{1}\) are orthogonal to \(\varphi_{n}\) for \(n>1\), we have the following.$$ \begin{aligned} \int_{a}^{b}(\alpha x+\beta) \varphi_{n}(x) d x &=\int_{a}^{b} a x \psi_{n}(x) d x+\int_{a}^{b} \beta \varphi_{n}(x) d x \\ &=\int_{a}^{b}\left(\square_{0}\right) \varphi_{1}(x) \varphi_{n}(x) d...
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Assess a subjective triangular probability distribution for the random variable, X, where X is defined as...
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function . What is the expected increase in your
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Let \(\left\{\varphi_{n}(x)\right\}\) be an orthogonal set of functions on \([a, b]\) such that \(\psi_{0}(x)=1\) and \(\varphi_{1}(x)=x\), Show that \(\int_{a}^{b}(\alpha x+\beta) \varphi_{n}(x) d x=0\) for \(n=2,3, \ldots\) and any constants \(\alpha\) and \(\beta\),First we note that \(\alpha x+\beta=(\square) \Phi_{1}(x)+(\square \quad) \Psi_{0}(x)\).Using this together with the fact that \(\varphi_{0}\) and \(\varphi_{1}\) are orthogonal to \(\varphi_{n}\) for \(n>1\), we have the following.$$ \begin{aligned} \int_{a}^{b}(\alpha x+\beta) \varphi_{n}(x) d x &=\int_{a}^{b} a x \psi_{n}(x) d x+\int_{a}^{b} \beta \varphi_{n}(x) d x \\ &=\int_{a}^{b}\left(\square_{0}\right) \varphi_{1}(x) \varphi_{n}(x) d...
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function . What is the expected increase in your
commute for the next snowstorm?
Estimate E[f(X)] by creating simple Monte Carlo
simulation in Excel.
Here are some equations for the triangular distribution that may
help you:...
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