Problems

Solution: The tangent to the curve \(y = f(x)\) at \(x = a\) has the equation \(\frac{y-f(a)}{x-a} = f'(a) = g(a)+(a-p)g'(a)\). This passes through \((p,0)\) iff \begin{align*} && \frac{-f(a)}{p-a} &= g(a)+(a-p)g'(a) \\ \Leftrightarrow && -f(a) &= (p-a)g(a) -(a-p)^2g'(a) \\ \Leftrightarrow && -f(a) &= -f(a) -(a-p)^2g'(a) \\ \Leftrightarrow && 0 &= g'(a) \\ \end{align*}

1. In this case \(g(x) = A(x-q)(x-r) = A(x^2-(q+r)x+qr)\) and so we must have that \(g'(a) = 0\), ie \(A(2a-(q+r)) = 0 \Rightarrow 2a = q+r\) The gradient is \(g(a) +(a-p)g'(a) = g(a) = A(a-q)(a-r)\)
2. By the same reasoning, but with \(g(x) = A(x-p)(x-q)\) we have the gradient is \(A(c-p)(c-r)\). This is parallel iff \begin{align*} && (c-p)(c-r) &= (a-q)(a-r) \end{align*} The tangent at \(x = q\) is \(\frac{y-0}{x-q} = A(q-p)(q-r)\) or \( y = A(q-p)(q-r)(x-q)\)

Solution:

1. Suppose \(n = 1\), then all polynomials are reflexive (since \(x - a_1\) has the root \(a_1\). Suppose \(n = 2\), then we want \begin{align*} && x^2-a_1x+a_2 &= (x-a_1)(x-a_2) \\ &&&= x^2-(a_1+a_2)x+a_1a_2 \\ \Rightarrow && a_2 &= 0 \\ \end{align*} So all polynomials of the form \(x^2-a_1x\) work and no others. Suppose \(n = 3\) then we want \begin{align*} && x^3-a_1x^2+a_2x-a_3 &= (x-a_1)(x-a_2)(x-a_3) \\ &&&= x^3-(a_1+a_2+a_3)x+(a_1a_2+a_1a_3+a_2a_3)x-a_1a_2a_3 \\ \Rightarrow && a_2+a_3 &= 0 \\ && a_2a_3 &= a_2 \\ \Rightarrow && -a_2^2 &= a_2 \\ \Rightarrow && a_2 &= 0, -1 \\ && -a_1a_2^2 &= -a_2 \\ \Rightarrow && a_2 &= 0, a_2 = 1/a_1 \end{align*} So we need either \(x^3-a_1x\) or \((x+1)^2(x-1) = x^3+x^2-x-1\)
2. Suppose \(n > 3\) then \begin{align*} && \sum a_i^2 &= \left (\sum a_i \right)^2 - 2 \sum_{i < j} a_i a_j \\ && &= a_1^2 - 2a_2 \\ \Rightarrow && 2a_2 &= a_1^2 - \sum a_i^2 \\ &&&= -a_2^2 - a_3^2 - \cdots - a_n^2 \end{align*} So \((a_2+1)^2 = 1-a_3^2 -\cdots -a_n^2\) so if \(a_n > 0\) (or any other \(a_i, i > 2\) for that matter) then we must have \(a_n = \pm 1, a_{3}, \ldots a_{n-1} = 0\), but if \(a_n = \pm 1\) \(x = 0\) is not a root. Therefore we must have \(a_0\) and \(a_i = 0\) for all \(i > 3\)
3. The only reflexive polynomials therefore must be \(x^n - kx^{n-1}\) and \(x^{n+3}+x^{n+2}-x^{n+1}-x^n\)

Solution:

1. \(y = e^x(2x^2-5x+2) = e^x(2x-1)(x-2)\), we also have \(y' = e^x(2x^2-5x+2 + 4x-5) = e^x(2x^2-x-3) = e^x(2x-3)(x+1)\) \(y(-1) = \frac{9}{e}\), \(y(\frac32) = -e^{3/2}\)
   

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   If \(k < -e^{3/2}\) there are no solutions. If \(k = -e^{3/2}\) there is a unique solution. If \(-e^{3/2} < k \leq 0\) there are two solutions. If \(0 < k < \frac{9}{e}\) there are three solutions. Otherwise there is a unique solution.
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Solution: \begin{align*} {\mathrm D}^2 x^a &= x\frac{\d\ }{\d x}\left( x\frac{\d}{\d x} \left ( x^a \right) \right) \\ &= x\frac{\d\ }{\d x}\left( ax^a \right) \\ &= a^2 x^a \end{align*}

1. Claim: \({\mathrm D^n}(x^a) =a^n x^a\) Proof: Induct on \(n\). Base cases we have already seen, so consider \(D^{k+1}(x^a) = D(a^k x^a) = a^{k+1}x^a\) as required. Claim: \({\mathrm D}\) is linear, ie \({\mathrm D}(f(x) + g(x)) = {\mathrm D}(f(x)) + {\mathrm D}(g(x))\) Proof: \begin{align*} {\mathrm D}(f(x) + g(x)) &= x\frac{\d\ }{\d x}\left(f(x) + g(x) \right) \\ &= x\frac{\d\ }{\d x}f(x) + x\frac{\d\ }{\d x}g(x) \\ &= {\mathrm D}(f(x)) + {\mathrm D}(g(x)) \end{align*} Claim: If \(p(x)\) is a polynomial degree \(r\) then \({\mathrm D}^n p(x)\) is a polynomial degree \(n\). Proof: Since \({\mathrm D}\) is linear, it suffices to prove this for a monomial of degree \(n\), but this was already proven in the first question.
2. Claim: If \(f(x)\) is some polynomial, \({\mathrm D}((1-x)^m f(x))\) is divisible by \((1-x)^{m-1}\) Proof: \({\mathrm D}((1-x)^mf(x)) = -xm(1-x)^{m-1}f(x) + (1-x)^mxf'(x) = x(1-x)^{m-1}((1-x)f'(x)-xf(x))\) as required. Therefore repeated application of \({\mathrm D}\) will reduce the factor of \(1-x\) by at most \(1\) each time as required.
3. \begin{align*} {\mathrm D}^n(1-x)^m &= {\mathrm D}^n \left ( \sum_{r=0}^m \binom{m}{r}(-1)^r x^r\right) \\ &= \sum_{r=0}^m {\mathrm D}^n \left ( \binom{m}{r}(-1)^r x^r \right ) \\ &= \sum_{r=0}^m\binom{m}{r}(-1)^r r^n x^r \end{align*} Since the left-hand side is divisible by \(1-x\), if we substitute \(x = 1\), the sum must be \(0\), i.e., we get the desired result.
4. On each application of \({\mathrm D}\) to \((1-x)^m f(x)\) we end up with a term in the form \(x(1-x)^{m-1}(x)\) and a term of the form \((1-x)^m\). After the latter term will be annihilated once we evaluate at \(x = 1\) because there will be insufficient applications to remove the factors of \(1-x\). Therefore we only need to focus on the term which does not get annihilated. This term is will be \((-x)^n n \cdot (n-1) \cdots 1\), so \(f_n(1) = (-1)^n n!\) as required. Alternatively: \begin{align*} {\mathrm D}^n((1-x)^n) &= D^{n-1}(-nx(1-x)^{n-1}) \\ &= -n{\mathrm D}^{n-1}(x(1-x)^{n-1}) \\ &= -n{\mathrm D}^{n-1}((x-1+1)(1-x)^{n-1}) \\ &= -n{\mathrm D}^{n-1}(-(1-x)^{n}+(1-x)^{n-1}) \\ &= -n{\mathrm D}^{n-1}(-(1-x)^{n})-n{\mathrm D}^{n-1}((1-x)^{n-1}) \\ \end{align*} Therefore, when this is evaluated at \(x = 1\), recursively, we will have \(f_n(1) = -nf_{n-1}(1)\), in particular, \(f_n(1) = (-1)^n n!\)

Solution:

1. \begin{align*} &&(n+1)^k &= \sum_{r=0}^n \left [ (r+1)^k - r^k \right] \\ &&&= \sum_{r=0}^n \left [ \left ( \binom{k}{0}r^k+\binom{k}1r^{k-1} + \binom{k}{2}r^{k-2} + \cdots + \binom{k}{k} 1 \right) - r^k\right] \\ &&&= \sum_{r=0}^n \left ( \binom{k}1r^{k-1} + \binom{k}{2}r^{k-2} + \cdots + \binom{k}{k} 1 \right) \\ &&&=k \sum_{r=0}^n r^{k-1} + \binom{k}{2}\sum_{r=0}^nr^{k-2} + \cdots + \binom{k}{k} \sum_{r=0}^n 1 \\ &&&= kS_{k-1}(n) + \binom{k}2 S_{k-2}(n) + \cdots +\binom{k}{k-1}S_1(n) + (n+1) \\ \Rightarrow && k S_{k-1}(n) &= (n+1)^k -(n+1) -\binom{k}2 S_{k-2}(n) - \cdots -\binom{k}{k-1}S_1(n) \\ && 4S_3(n) &= (n+1)^4-(n+1) - \binom{4}{2} \frac{n(n+1)(2n+1)}{6} - \binom{4}{3} \frac{n(n+1)}{2} \\ &&&= (n+1) \left ( (n+1)^3-1 - n(2n+1)-2n \right) \\ &&&= (n+1) \left ( n^3+3n^2+3n+1-1 - 2n^2-3n \right) \\ &&&= (n+1) \left ( n^3+n^2 \right) \\ &&&= n^2(n+1)^2 \\ \Rightarrow && S_3(n) &= \frac{n^2(n+1)^2}{4} \\ \\ &&5S_4(n) &=(n+1)^5-(n+1) - \binom{5}{2} \frac{n^2(n+1)^2}4 - \binom{5}{3} \frac{n(n+1)(2n+1)}{6} - \binom{5}{4} \frac{n(n+1)}{2} \\ &&&= (n+1) \left ((n+1)^4 - 1-\frac{5n^2(n+1)}{2} - \frac{5n(2n+1)}{3} -\frac{5n}{2}\right)\\ &&&= \frac{n+1}{6} \left (6(n+1)^4-6-15n^2(n+1)-10n(2n+1)-15n \right) \\ &&&= \frac{n+1}{6} \left (6n^4+24n^3+36n^2+24n+6 -6-15n^3-15n^2-20n^2-10n-15n\right) \\ &&&= \frac{n+1}{6} \left (6n^4+9n^3+n^2-n\right) \\ &&&= \frac{(n+1)n(2n+1)(3n^2+3n-1)}{6} \\ \Rightarrow && S_4(n) &= \frac{(n+1)n(2n+1)(3n^2+3n-1)}{30} \end{align*}
2. Proceeding by induction, since \(S_k(n)\) is a polynomial of degree \(k+1\) for small \(k\), we can see that \[ (k+1)S_k(n) = \underbrace{(n+1)^{k+1}}_{\text{poly deg }=k+1} - \underbrace{(n+1)}_{\text{poly deg}=1} - \underbrace{\binom{k+1}{2}S_{k-1}(n)}_{\text{poly deg}=k} - \underbrace{\cdots}_{\text{polys deg}< k} - \underbrace{\binom{k+1}{k} S_1(n)}_{\text{poly deg}=1}\] therefore \(S_k(n)\) is a polynomial of degree \(k+1\) (in fact with leading coefficient \(\frac{1}{k+1}\). Since \(S_k(0) = \sum_{r=0}^{0} r^k = 0\) there is no constant term, and since \(S_k(1) = \sum_{r=0}^1 r^k = 1\) the sum of the coefficients is \(1\)

Solution:

1. \(\,\) \begin{align*} && y &= 2x^3-bx^2+cx \\ \Rightarrow && y' &= 6x^2-2bx+c \end{align*} We must have \(p, q\) are the roots of this equation, ie \(\frac13b = p+q, \frac16c = pq\)
2. The point of inflection will be at \(\frac{b}6\) The graph will have rotational symmetry of \(180^{\circ}\) about the point of inflection.
   

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3. \begin{align*} && m-n &= 2(p^3-q^3)-b(p^2-q^2)+c(p-q) \\ &&&= (p-q)(2(p^2+qp+q^2)-b(p+q)+c) \\ &&&= (p-q)(2(p^2+qp+q^2)-3(p+q)^2+6pq) \\ &&&= (p-q)(-p^2-q^2+2pq) \\ &&&= (q-p)^3 \end{align*}
4. The area of \(R\) is \begin{align*} A &= \frac12 bh \\ &= \frac12 (q-p)(m-n) = \frac12(q-p)^4 \end{align*} as required.

Solution: \begin{align*} && y &= x^2e^{-x^2} \\ \Rightarrow && y' &= 2xe^{-x^2} +x^2 \cdot (-2x)e^{-x^2} \\ &&&= e^{-x^2}(2x-2x^3) \\ &&&= 2e^{-x^2}x(1-x^2) \end{align*} Therefore the derivative is zero iff \(x = 0, \pm 1\) \begin{align*} && y &= \P(x) e^{-x^2} \\ \Rightarrow && y' &= e^{-x^2} (\P'(x)-2x\P(x)) \end{align*} Therefore we want \(\P'(x) - 2x\P(x) = Kx(x^2-a^2)(x^2-b^2)\) Since this has degree \(5\), we should look at polynomials degree \(4\) for \(\P\). We can also immediately see that \(0\) is a root of \(\P'(x)\), so \(\P(x) = a_4x^4+a_3x^3+a_2x^2+a_0\). WLOG \(a_4 = 1\) and \(K = -2\), so \begin{align*} && -2(x^5-(a^2+b^2)x^3+a^2b^2x) &= 4x^3+3a_3x^2+2a_2x- 2x(x^4+a_3x^3+a_2x^2+a_0) \\ &&&= -2x^5-2a_3 x^4+(4-2a_2)x^3+(2a_2-2a_0)x \\ \Rightarrow && a_3 &= 0 \\ && a^2+b^2 &= 2-a_2 \\ \Rightarrow && a_2 &= 2-a^2-b^2 \\ && a^2b^2 &= a_0-a_2 \\ \Rightarrow && a_0 &= a^2b^2 + 2-a^2-b^2 \\ \Rightarrow && \P(x) &= x^4+(2-a^2-b^2)x^2+(a^2-1)(b^2-1)x \end{align*}

Solution: \begin{align*} && y &= x(x+1)(x-2)^4 \\ \Rightarrow && y' &= (x+1)(x-2)^4+x(x-2)^4+4x(x+1)(x-2)^3 \\ &&&= (x-2)^3 \left ( (2x+1)(x-2)+4x(x+1) \right) \\ &&&= (x-2)^3 \left (2x^2-3x-2+4x^2+4x \right) \\ &&&=(x-2)^3(6x^2+x-2) \\ &&&=(x-2)^3(2x-1)(3x+2) \end{align*} Therefore there are stationary points at \((2,0), (\frac12, -\frac{625}{64}), (-\frac23, -\frac{4078}{81})\) \((0,2)\) is a minimum by considering the sign of \(y'\) either side. \( (-\frac23, \frac{2560}{729})\) is a minimum, since it's the first stationary point. \( (\frac12, \frac{243}{64})\) is a maximum since you can't have consecutive minima and the second derivative is clearly non-zero.

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Solution: \begin{align*} n = -1: && (-1)^2 &= s - r+q -p \\ n = 0: && 1 + 0 &= s \\ n = 1: && 1 + 1 &= s + r + q + p \\ n = 2: && 2 + 2^2 &= s + 2r + 4q + 8p \\ \Rightarrow &&& \begin{cases} 1 &= s \\ 1 &= s - r + q - p \\ 2 &= s + r + q + p \\ 6 &= s + 2r + 4q + 8p \end{cases} \\ \Rightarrow && s &= 1 \\ && q &= \frac12 \\ &&& \begin{cases} \frac12 &= r + p \\ 3 &= 2r + 8p \end{cases} \\ \Rightarrow && r &= \frac16 \\ && p &= \frac13 \\ \Rightarrow && \sum_{r=0}^n r^2 &= 1 + \frac16 n + \frac12 n^2 + \frac13 n^3 - (-1)^2 \\ &&&= \frac{n}{6} \l 1 + 3n + 2n^2 \r \\ &&&= \frac{n(n+1)(2n+1)}{6} \end{align*} Similarly, \begin{align*} n = -2: && (-2)^3 &= e - 2d + 4c - 8b + 16a \\ n = -1: && -8 + (-1)^3 &= e -d+c-b+a \\ n = 0: && -9 + 0^3 &= e \\ n = 1: && -9 + 1^3 &= e+d+c+b+a \\ n = 2: && -8 + 2^3 &= e+2d+4c+8b+16a \\ \Rightarrow &&& \begin{cases} -9 &= e \\ -9 &= e - d+c -b + a \\ -8 &= e +d+c+b+a \\ -8 &= e-2d+4c-8b+16a \\ 0 &= e+2d+4c+8b+16a \\ \end{cases} \\ \Rightarrow && e &= -9 \\ \Rightarrow &&& \begin{cases} 1 &= 2c+2a \\ 10 &= 8c+32a \\ 1 &= 2d+2b \\ 8 &= 4d+16b \\ \end{cases} \\ \Rightarrow && a &= \frac14 \\ && c &= \frac14 \\ && b &= \frac12 \\ && d &= 0 \\ \\ \Rightarrow && \sum_{r=0}^n r^3 &= -9 + \frac14n^2 + \frac12 n^3+\frac14 n^4 -((-1)^3+(-2)^3) \\ &&&= \frac14n^2 \l1 + 2n+n^2\r \\ &&&= \frac{n^2(n+1)^2}{4} \end{align*} as required

Solution: \begin{align*} && f'(x) &= mx^{m-1}(x-1)^n + nx^m(x-1)^{n-1} \\ &&&= (m(x-1)+nx)x^{m-1}(x-1)^{n-1} \\ &&&= (x(m+n) - m)x^{m-1}(x-1)^{n-1} \\ \end{align*} Therefore when \(x = \frac{m}{m+n}\) there is a stationary point with \(0 < x < 1\). \begin{align*} && f''(x) &= m(m-1)x^{m-2}(x-1)^n + 2mnx^{m-1}(x-1)^{n-1} + n(n-1)x^{m}(x-1)^{n-2} \\ &&&= (m(m-1)(x-1)^2 +2mnx(x-1)+n(n-1)x^2)x^{m-2}(1-x)^{n-2} \\ \Rightarrow && f'' \left ( \frac{m}{m+n} \right) &= \left ( m(m-1) \frac{n^2}{(m+n)^2} - 2mn\frac{mn}{(m+n)^2} + n(n-1) \frac{m^2}{(m+n)^2} \right) \frac{m^{m-2}}{(m+n)^{m-2}} \frac{(-1)^{n-2}n^{n-2}}{(m+n)^{n-2}} \\ &&&= (-1)^{n-2}\frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-2}} \left ( (m-1)n-2mn+(n-1)m\right) \\ &&&= (-1)^{n-2}\frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-2}} \left ( -m-n\right) \\ &&&= (-1)^{n-1} \frac{m^{m-1}n^{n-1}}{(m+n)^{m+n-3}} \end{align*} Therefore this is positive (and a minimum) when \(n\) is odd and negative (and a maximum) when \(n\) is even.

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Solution: \begin{align*} && S_2(x) &= e^{x^3} \frac{d^2}{\d x^2} \left [e^{-x^3} \right] \\ &&&= e^{x^3} \frac{d}{\d x} \left [e^{-x^3}(-3x^2) \right] \\ &&&= e^{x^3} \left [e^{-x^3}(9x^4-6x) \right] \\ &&&=9x^4-6x \\ \\ && S_3(x) &= e^{x^3} \frac{\d^3}{\d x^3} \left [ e^{-x^3} \right]\\ &&&= e^{x^3} \frac{\d}{\d x} \left [ e^{-x^3}(9x^4-6x) \right ] \\ &&&= e^{x^3} e^{-x^3}\left [ (-3x^2)(9x^4-6x)+(36x^3-6) \right ] \\ &&&= -27x^6 +54x^3-6 \end{align*} Claim: \(S_n\) is a polynomial of degree \(2n\) with leading coefficient \((-3)^n\). Proof: Clearly this is true for \(n = 1, 2, 3\) by demonstration. Suppose it is true for some \(n = k\), then \begin{align*} && S_k(x) &= e^{x^2} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] \\ && (-3)^kx^{2k} +\cdots &= e^{x^3} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] \\ \Rightarrow && \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] &= e^{-x^3} \left ( (-3)^kx^{2k} +\cdots\right) \\ \Rightarrow && \frac{\d^k}{\d x^k}\left [ e^{x^3}\right] &= e^{-x^3} (-3x^2)\left ( (-3)^kx^{2k} +\cdots\right) + e^{-x^3} S_k'(x) \\ &&&= e^{-x^3} \left (\underbrace{ (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x)}_{\deg =2k+2}\right) \\ \Rightarrow && S_{k+1}(x) &= (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x) \end{align*} And therefore \(S_{k+1}\) is a polynomial degree \(2(k+1)\) with leading coefficient \((-3)^{k+1}\) so by induction it's true for all \(n\). If \(S'_n(a) = 0\) then \(S_{n+1}(a) = (-3a^2)S_n(a) + S_n'(a) \Rightarrow S_{n+1}(a)S_n(a) = -3 (aS_n(a))^2 \leq 0\)

Solution: Since this is true for all cubic polynomials, it must be true in particular for \(1, x, x^2, x^3\), therefore: \begin{align*} \int_{-1}^{1} 1 {\mathrm d}t &=a_{1}+a_{2} &=2\\ \int_{-1}^{1} x {\mathrm d}t &=a_{1}u_1+a_{2}u_2 &= 0 \\ \int_{-1}^{1} x^2 {\mathrm d}t &=a_{1}u_1^2+a_{2}u_2^2 &= \frac23\\ \int_{-1}^{1} x^3 {\mathrm d}t &=a_{1}u_1^3+a_{2}u_2^3 &= 0\\ \end{align*} \begin{align*} && \begin{cases} a_{1}+a_{2} &=2 \\ a_{1}u_1+a_{2}u_2 &= 0 \\ a_{1}u_1^2+a_{2}u_2^2 &= \frac23\\ a_{1}u_1^3+a_{2}u_2^3 &= 0\\ \end{cases} \\ \Rightarrow && \begin{cases} a_{1}(u_1^2 - \frac13) + a_{2}(u_2^2 - \frac13) &= 0 \\ a_{1}u_1(u_1^2 - \frac13) + a_{2}u_2(u_2^2 - \frac13) &= 0 \end{cases} \\ \Rightarrow && \begin{cases} u_i = \pm \frac1{\sqrt{3}} \\ a_i = 1\end{cases} \end{align*} Therefore we have: \[\int_{-1}^{1}{\mathrm P}(t)\,{\mathrm d}t ={\mathrm P} \l \frac1{\sqrt{3}} \r+{\mathrm P}\l -\frac1{\sqrt{3}} \r \] [Note: this question is actually asking about Gauss-Legendre polynomials, and could be done directly by appealing to standard results]

Solution: \begin{align*} f(x) & =nx-\binom{n}{2}\frac{x^{2}}{2}+\binom{n}{3}\frac{x^{3}}{3}-\cdots+(-1)^{r+1}\binom{n}{r}\frac{x^{r}}{r}+\cdots+(-1)^{n+1}\frac{x^{n}}{n} \\ f'(x) &= n - \binom{n}{2} x + \binom{n}{3}x^2 - \cdots (-1)^{r+1} \binom{n}{r} + \cdots + (-1)^{n+1} x^{n-1} \\ &= \frac{1-(1-x)^n}{x} \end{align*} Therefore, since \(\displaystyle f(x) = \int_0^xf'(t)\,dt\) \begin{align*} f(x) &= \int_0^x \frac{1 - (1-t)^n}{t} \, dt \\ &= \int_{1}^{1-x} \frac{1-y^n}{1-y} (-1)\, dy \tag{Let \(y = 1-t, \frac{dy}{dt} = -1\)} \\ &= \boxed{\int_{1-x}^1 \frac{1-y^n}{1-y} dy} \\ &= \int_{1-x}^1 \l 1 + y + y^2 + \cdots + y^{n-1} \r \, dy \\ &= \left [ y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots + \frac{y^n}{n} \right]_{1-x}^1 \\ \end{align*} So when \(x = 1, 1-x = 0\) so we exactly have the sum required.