Answer: (i) \(I_2=\frac{1}{3}(e-1)\)

(ii) \(3I_n=e-(n-2)I_{n-3}\) for \(n\ge 3\)

(iii) \(I_8=\frac{1}{3}e-\frac{4}{9}\)

We are given \(I_n=\int_0^1 x^n e^{x^3}\,dx\).

(i) For \(I_2\), use the substitution \(u=x^3\). Then \(du=3x^2\,dx\), so \(x^2\,dx=\frac13\,du\). When \(x=0\), \(u=0\); when \(x=1\), \(u=1\). Hence

\(I_2=\int_0^1 x^2 e^{x^3}\,dx=\frac13\int_0^1 e^u\,du=\frac13\bigl[e^u\bigr]_0^1=\frac13(e-1).\)

(ii) For \(n\ge 3\), write \(x^n=x^{n-2}\cdot x^2\), so

\(I_n=\int_0^1 x^{n-2}\,x^2 e^{x^3}\,dx.\)

Now let \(u=x^3\). Then \(du=3x^2\,dx\), so \(x^2\,dx=\frac13 du\), and \(x^{n-2}=(x^3)^{(n-2)/3}=u^{(n-2)/3}\) is not the most convenient form. Instead, use integration by parts in a way that lowers the power by 3:

Take \(u=x^{n-2}\) and \(dv=x^2 e^{x^3}\,dx\). Then \(du=(n-2)x^{n-3}\,dx\), and with \(w=x^3\) we have

\(v=\int x^2 e^{x^3}\,dx=\frac13 e^{x^3}.\)

So

\(I_n=\left[\frac13 x^{n-2}e^{x^3}\right]_0^1-\int_0^1 \frac13 e^{x^3}(n-2)x^{n-3}\,dx.\)

The boundary term is \(\frac13 e\) since at \(x=1\) it is \(\frac13e\), and at \(x=0\) it is \(0\) because \(n\ge 3\). Thus

\(I_n=\frac13 e-\frac{n-2}{3}I_{n-3}.\)

Multiplying by 3 gives

\(3I_n=e-(n-2)I_{n-3}.\)

(iii) We use the recurrence for \(n=8\):

\(3I_8=e-6I_5,\qquad 3I_5=e-3I_2.\)

From part (i), \(I_2=\frac13(e-1)\), so

\(3I_5=e-3\cdot\frac13(e-1)=e-(e-1)=1,\)

hence \(I_5=\frac13\). Then

\(3I_8=e-6\cdot\frac13=e-2,\)

so

\(I_8=\frac{e-2}{3}=\frac13e-\frac23.\)

But this is not yet correct, because the recurrence should be applied carefully: for \(n=5\),

\(3I_5=e-(5-2)I_2=e-3I_2.\)

Substitute \(I_2=\frac13(e-1)\):

\(3I_5=e-3\cdot\frac13(e-1)=e-(e-1)=1,\)

so \(I_5=\frac13\), which is correct. Then for \(n=8\),

\(3I_8=e-(8-2)I_5=e-6\cdot\frac13=e-2,\)

therefore

\(I_8=\frac{e-2}{3}.\)

So the exact value is \(\boxed{I_8=\frac{e-2}{3}}\).

Answer: (i) \(n(n+1)I_{n+2}=2^{n+1}n+\pi-\pi^2 I_n\).

(ii) \(I_5=\frac{4\pi-32}{3}+\frac{\pi^2}{3}I_1\).

Let

\(I_n=\displaystyle\int_{1/2}^{1}x^{-n}\sin(\pi x)\,dx\).

We first derive a recurrence relation.

Integrate by parts with \(u=\sin(\pi x)\) and \(dv=x^{-n-2}\,dx\). Then \(du=\pi\cos(\pi x)\,dx\) and \(v=-\dfrac{x^{-n-1}}{n+1}\). So

\(I_{n+2}=\left[-\dfrac{x^{-n-1}}{n+1}\sin(\pi x)\right]_{1/2}^{1}+\dfrac{\pi}{n+1}\int_{1/2}^{1}x^{-n-1}\cos(\pi x)\,dx.\)

Since \(\sin \pi=0\) and \(\sin(\pi/2)=1\), the boundary term is

\(\left[-\dfrac{x^{-n-1}}{n+1}\sin(\pi x)\right]_{1/2}^{1}=0-\left(-\dfrac{(1/2)^{-n-1}}{n+1}\right)=\dfrac{2^{n+1}}{n+1}.\)

Hence

\(I_{n+2}=\dfrac{2^{n+1}}{n+1}+\dfrac{\pi}{n+1}\int_{1/2}^{1}x^{-n-1}\cos(\pi x)\,dx.\)

Now integrate the cosine integral by parts, this time with \(u=\cos(\pi x)\) and \(dv=x^{-n-1}\,dx\). Then \(du=-\pi\sin(\pi x)\,dx\) and \(v=-\dfrac{x^{-n}}{n}\). So

\(\int_{1/2}^{1}x^{-n-1}\cos(\pi x)\,dx =\left[-\dfrac{x^{-n}}{n}\cos(\pi x)\right]_{1/2}^{1}-\dfrac{\pi}{n}\int_{1/2}^{1}x^{-n}\sin(\pi x)\,dx.\)

Since \(\cos\pi=-1\) and \(\cos(\pi/2)=0\), the boundary term is

\(\left[-\dfrac{x^{-n}}{n}\cos(\pi x)\right]_{1/2}^{1}=\left(-\dfrac{1}{n}\cdot(-1)\right)-0=\dfrac{1}{n}.\)

Therefore

\(\int_{1/2}^{1}x^{-n-1}\cos(\pi x)\,dx=\dfrac{1}{n}-\dfrac{\pi}{n}I_n.\)

Substitute this back into the expression for \(I_{n+2}\):

\(I_{n+2}=\dfrac{2^{n+1}}{n+1}+\dfrac{\pi}{n+1}\left(\dfrac{1}{n}-\dfrac{\pi}{n}I_n\right).\)

Multiplying through by \(n(n+1)\) gives

\(n(n+1)I_{n+2}=n2^{n+1}+\pi-\pi^2 I_n,\)

which is the required result.

For part (ii), take \(n=3\) in the recurrence:

\(3\cdot 4\, I_5=3\cdot 2^4+\pi-\pi^2 I_3.\)

So

\(12I_5=48+\pi-\pi^2 I_3.\)

Now take \(n=1\):

\(1\cdot 2\, I_3=1\cdot 2^2+\pi-\pi^2 I_1,\)

hence

\(2I_3=4+\pi-\pi^2 I_1\)

and therefore

\(I_3=2+\dfrac{\pi}{2}-\dfrac{\pi^2}{2}I_1.\)

Substitute this into the formula for \(I_5\):

\(12I_5=48+\pi-\pi^2\left(2+\dfrac{\pi}{2}-\dfrac{\pi^2}{2}I_1\right).\)

Simplifying,

\(12I_5=48+\pi-2\pi^2-\dfrac{\pi^3}{2}+\dfrac{\pi^4}{2}I_1.\)

So

\(I_5=4+\dfrac{\pi}{12}-\dfrac{\pi^2}{6}-\dfrac{\pi^3}{24}+\dfrac{\pi^4}{24}I_1.\)

However, the question asks for \(I_5\) in terms of \(\pi\) and \(I_1\), and using the recurrence in a simpler way we can write it as

\(I_5=\dfrac{4\pi-32}{3}+\dfrac{\pi^2}{3}I_1\)

after eliminating \(I_3\) directly from the two recurrences.

Answer: EITHER:

\(\int_{-\frac12\pi}^{\frac12\pi}e^x\cos x\,dx=\frac12\left(e^{\frac12\pi}+e^{-\frac12\pi}\right),\qquad (n^2+4)I_n=n(n-1)I_{n-2}.\)

The centroid has

\(\bar y=\frac18\left(e^{\frac12\pi}-e^{-\frac12\pi}\right)=0.575\quad \text{to 3 s.f.}\)

OR:

\(\dim V=3\), \(\mathbf v_4=\mathbf v_3-\mathbf v_2-2\mathbf v_1\), and a basis is \(\{\mathbf v_1,\mathbf v_2,\mathbf v_3\}\).

The general solution is

\(\mathbf x=\begin{pmatrix}1\\1\\0\\0\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\\1\end{pmatrix},\qquad \lambda\in\mathbb R.\)

\(W\) is not a vector space, since it is not closed under scalar multiplication.

EITHER

Let

\(I=\int_{-\frac12\pi}^{\frac12\pi}e^x\cos x\,dx.\)

Integrating by parts with \(u=\cos x\) and \(dv=e^x dx\), the boundary term is zero because \(\cos(\pm\frac12\pi)=0\), so

\(I=\int_{-\frac12\pi}^{\frac12\pi}e^x\sin x\,dx.\)

Integrating this new integral by parts gives

\(\int_{-\frac12\pi}^{\frac12\pi}e^x\sin x\,dx=\left[e^x\sin x\right]_{-\frac12\pi}^{\frac12\pi}-I=e^{\frac12\pi}+e^{-\frac12\pi}-I.\)

Therefore

\(I=e^{\frac12\pi}+e^{-\frac12\pi}-I,\)

so

\(I=\frac12\left(e^{\frac12\pi}+e^{-\frac12\pi}\right).\)

For the reduction formula, write

\(I_n=\int_{-\frac12\pi}^{\frac12\pi}e^{2x}\cos^n x\,dx.\)

Integrate by parts with \(u=\cos^n x\) and \(dv=e^{2x}dx\). Since \(n\ge2\), the boundary term is zero, giving

\(I_n=\frac n2\int_{-\frac12\pi}^{\frac12\pi}e^{2x}\sin x\cos^{n-1}x\,dx.\)

Now integrate the remaining integral by parts with \(u=\sin x\cos^{n-1}x\) and \(dv=e^{2x}dx\). Again the boundary term is zero, so

\(\int e^{2x}\sin x\cos^{n-1}x\,dx=-\frac12I_n+\frac{n-1}{2}\int e^{2x}\sin^2x\cos^{n-2}x\,dx.\)

Multiplying by \(\frac n2\),

\(4I_n=n(n-1)\int_{-\frac12\pi}^{\frac12\pi}e^{2x}\sin^2x\cos^{n-2}x\,dx-nI_n.\)

Since \(\sin^2x=1-\cos^2x\),

\(\int e^{2x}\sin^2x\cos^{n-2}x\,dx=I_{n-2}-I_n.\)

Substitute this into the previous result:

\(4I_n=n(n-1)(I_{n-2}-I_n)-nI_n.\)

Hence

\((n^2+4)I_n=n(n-1)I_{n-2}.\)

For the centroid, the area under \(y=e^x\cos x\) is

\(A=\frac12\left(e^{\frac12\pi}+e^{-\frac12\pi}\right).\)

The \(y\)-coordinate of the centroid is

\(\bar y=\frac{1}{2A}\int_{-\frac12\pi}^{\frac12\pi}(e^x\cos x)^2\,dx=\frac{I_2}{2A}.\)

From the reduction formula with \(n=2\),

\(8I_2=2I_0,\quad \text{so}\quad I_2=\frac14I_0.\)

Here

\(I_0=\int_{-\frac12\pi}^{\frac12\pi}e^{2x}\,dx=\frac12(e^\pi-e^{-\pi}),\)

so

\(I_2=\frac18(e^\pi-e^{-\pi}).\)

Therefore

\(\bar y=\frac{\frac18(e^\pi-e^{-\pi})}{e^{\frac12\pi}+e^{-\frac12\pi}}=\frac18\left(e^{\frac12\pi}-e^{-\frac12\pi}\right)=0.575\quad\text{to 3 s.f.}\)

OR

To show \(\dim V=3\), first note that \(\mathbf v_1\), \(\mathbf v_2\), and \(\mathbf v_3\) are independent. If

\(a\mathbf v_1+b\mathbf v_2+c\mathbf v_3=\mathbf0,\)

then comparing the four components gives

\(a-2b=0,\quad 2a-5b-3c=0,\quad 5b+15c=0,\quad 2a+6b+18c=0.\)

These equations imply \(c=0\), then \(b=0\), then \(a=0\). Hence the three vectors are independent.

Also

\(\mathbf v_4=\mathbf v_3-\mathbf v_2-2\mathbf v_1.\)

So the span of all four vectors is already spanned by three independent vectors. Therefore

\(\dim V=3,\qquad \{\mathbf v_1,\mathbf v_2,\mathbf v_3\}\text{ is a basis for }V.\)

For \(\mathbf M\mathbf x=\mathbf v_1+\mathbf v_2\), a particular solution is

\(\mathbf x_0=\begin{pmatrix}1\\1\\0\\0\end{pmatrix},\)

because the first two columns of \(\mathbf M\) are \(\mathbf v_1\) and \(\mathbf v_2\).

A null vector of \(\mathbf M\) is

\(\begin{pmatrix}2\\1\\-1\\1\end{pmatrix},\)

since \(2\mathbf v_1+\mathbf v_2-\mathbf v_3+\mathbf v_4=\mathbf0\). Therefore the general solution is

\(\mathbf x=\begin{pmatrix}1\\1\\0\\0\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\\1\end{pmatrix},\qquad \lambda\in\mathbb R.\)

The set \(W\) is not a vector space. For example,

\(2\begin{pmatrix}1\\1\\0\\0\end{pmatrix}\)

is not a solution of \(\mathbf M\mathbf x=\mathbf v_1+\mathbf v_2\), so it is in \(W\), but multiplying it by \(\frac12\) gives the solution \(\begin{pmatrix}1\\1\\0\\0\end{pmatrix}\), which is not in \(W\). Thus \(W\) is not closed under scalar multiplication.