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Use integration by parts in $I_{n+1}$, writing the integrand as $x^{2n+1}(x{e}^{-x^2})$. Since an antiderivative of $x{e}^{-x^2}$ is $-e^{-x^2}/2$,The boundary term vanishes because the Gaussian exponential function decays faster than any fixed power. These improper integrals are absolutely convergent, justifying the limiting integration by parts.

Starting from the permitted Gaussian integral $I_0=\sqrt{\pi }$, the Gaussian moment recurrence gives $I_1=\sqrt{\pi }/2$ and thenMore generally, iteration gives $I_n=(2n-1)!!\sqrt{\pi }/2^n$, using the double factorial with $(-1)!!=1$.

Assume that wrong-note events are independent events and that each note has the same error probability $p$, with $0\le p\le 1$. Then their count $N$ has a binomial distribution. Choose the $n$ mistaken positions in $\left(\genfrac{}{}{0ex}{}{100}{n}\right)$ ways and multiply the probabilities of the errors and correct notes:The endpoint cases $p=0$ or $p=1$ are understood by continuity, giving deterministic counts. More than half means at least $51$ mistakes, soThe single-note error probability alone would not determine this count distribution if mistakes were correlated; independence is the additional standard modeling assumption.

A uniformly shuffled deck makes all unordered five-card hands equally likely, giving $\left(\genfrac{}{}{0ex}{}{52}{5}\right)$ possible hands. Choose the rank appearing four times in $13$ ways; all four suits of that rank are then fixed. The remaining card can be any of the other $48$ cards. No hand is counted twice, because five cards cannot contain two different complete four-card ranks. This use of combinatorics gives the probabilityHere $\left(\genfrac{}{}{0ex}{}{52}{5}\right)=2598960$ is a binomial coefficient; the probability also simplifies to $1/4165\approx 0.000240096$.

For $f(x)=x^{1/3}$, the required derivatives areAt $a=8$ these give $f(8)=2$, $f^{\prime }(8)=1/12$, and $f^{\prime \prime }(8)=-1/144$. Setting $x-a=1$, the three-term Taylor polynomial yieldsBy the Taylor theorem with Lagrange remainder, the signed error isIt is positive, and ${\xi }^{-8/3}<8^{-8/3}=1/256$. ConsequentlyThe last comparison is exact because $5\cdot 4000=20000<20736$. Thus the estimate is an underestimate with the required rigorous absolute error bound.

With the first $n$ terms counted from the constant term, Taylor's theorem giveswhere $\xi$ lies between $a$ and $x$. Continuous derivatives through order $n$ on the joining interval are a sufficient hypothesis for this Taylor theorem with Lagrange remainder. The Taylor remainder measures the error of the finite Taylor polynomial; convergence of the corresponding infinite Taylor series is a separate question.

Since $1+x^2>0$ for all real $x$, divide the equation by that factor. Its linear coefficient can be integrated by polynomial division:Thus a positive integrating factor is $M=e^{3x^2/2}(1+x^2{)}^{-3/2}$. Multiplying the divided equation gives the product rule formThe integration by substitution $u=1+x^2$ givesThe initial condition $y(0)=1$ fixes $C-2=1$. The required solution is thereforeNo real singularity was introduced by the division or the integrating factor, so the solution extends over the entire real line.

For the first-order linear ordinary differential equation, seek an integrating factor $M(x)$ that makes its left side a product rule derivative. Multiplying by $M$ gives $M{y}^{\prime }+MPy$, and comparison with $(My{)}^{\prime }=M{y}^{\prime }+M^{\prime }y$ requires $M^{\prime }=PM$. On an interval where $P$ is continuous chooseThis factor is never zero. The transformed equation is $(My{)}^{\prime }=MQ$, so integration from $x_0$ to $x$ yieldsHere $C=y(x_0)$ for an initial value problem. The indefinite integral version is $y=e^{-\int Pdx}[C+\int Q{e}^{\int Pdx}dx]$, with any constant choices inside the integrating factor absorbed into $C$. Continuity of $P,Q$ is a convenient sufficient hypothesis for a classical differentiable solution of a differential equation.

The homogeneous characteristic equation is $r^2+2r=r(r+2)=0$, giving the complementary solution $C_1+C_2{e}^{-2x}$. For a particular solution, try $A{x}^2+Bx$. Substitution gives $2A+4Ax+2B=4x$, so $A=1$ and $B=-1$. Therefore the general solution isAs a direct check, the polynomial part contributes $2+2(2x-1)=4x$, while both complementary terms contribute zero.