Find $\int \frac{x}{\sqrt{2x-1}}\,dx.$ Using the substitution $t = \tan x$, find $\int \frac{1}{4\cos^{2}x + 9\sin^{2}x}\,dx.$ The diagram above shows part of the graph of $y = x^{2} + 3x$, with rectangles approximating the area under the curve from $x = 0$ to $x = 1$. The area under the curve may be approximated by the total area, $A$, of $(n-1)$ rectangles each of width $\tfrac{1}{n}$. Given that $\sum_{r=1}^{n-1} r^{2} = \tfrac{1}{6} n(n+1)(2n+1),$ show that $A = \frac{(n-1)(1ln-1)}{6n^{2}}$. Explain briefly how the value of $\int_{0}^{1} x^{2} + 3x\,dx$ can be deduced from this expression, and hence find this value exactly without integration.

1. Rewrite the numerator in terms of the expression in the denominator so that it can be split into integrable power terms.

$\int \frac{x}{\sqrt{2x-1}} \, dx = \int \frac{\frac{1}{2}(2x-1) + \frac{1}{2}}{\sqrt{2x-1}} \, dx$
$= \int \left[ \frac{1}{2}(2x-1)^{\frac{1}{2}} + \frac{1}{2}(2x-1)^{-\frac{1}{2}} \right] \, dx$

2. Integrate each term using the standard formula $\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)}$.

$= \frac{1}{2} \left[ \frac{(2x-1)^{\frac{3}{2}}}{\frac{3}{2} \cdot 2} \right] + \frac{1}{2} \left[ \frac{(2x-1)^{\frac{1}{2}}}{\frac{1}{2} \cdot 2} \right] + C$
$= \frac{1}{6}(2x-1)^{\frac{3}{2}} + \frac{1}{2}(2x-1)^{\frac{1}{2}} + C$

3. Factorise and simplify the integrated expression into a single term.

$= \frac{1}{6}(2x-1)^{\frac{1}{2}} \left[ (2x-1) + 3 \right] + C$
$= \frac{1}{6}\sqrt{2x-1}(2x+2) + C$
$= \frac{1}{3}(x+1)\sqrt{2x-1} + C$

1. Relate the integral's terms to the substitution $t = \tan x$ by applying algebraic manipulation.

Let $t = \tan x$. Differentiating with respect to $x$ gives:
$\frac{dt}{dx} = \sec^2 x \implies dt = \sec^2 x \,dx$
To introduce these terms into the integral, divide the numerator and the denominator by $\cos^2 x$:
$\int \frac{1}{4\cos^{2}x + 9\sin^{2}x}\,dx = \int \frac{\frac{1}{\cos^2 x}}{4 + 9\frac{\sin^2 x}{\cos^2 x}}\,dx = \int \frac{\sec^2 x}{4 + 9\tan^2 x}\,dx$

2. Express the integral entirely in terms of the new variable $t$.

Substitute $t = \tan x$ and $dt = \sec^2 x \,dx$ into the integral:
$\int \frac{\sec^2 x}{4 + 9\tan^2 x}\,dx = \int \frac{1}{4 + 9t^2}\,dt$

3. Evaluate the integral in terms of $t$ using the standard integration formula for inverse tangent.

Using the standard result $\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$:
$\int \frac{1}{2^2 + (3t)^2}\,dt = \frac{1}{3} \left[ \frac{1}{2} \tan^{-1}\left(\frac{3t}{2}\right) \right] + C$
$= \frac{1}{6} \tan^{-1}\left(\frac{3t}{2}\right) + C$

4. Express the final answer in terms of the original variable $x$.

Replace $t$ with $\tan x$ to obtain the final answer:
$\frac{1}{6} \tan^{-1}\left(\frac{3\tan x}{2}\right) + C$

1. Set up the total rectangle area as a sum

Each rectangle has width $\tfrac1n$ and height equal to the curve value at the right endpoint $x=\tfrac{r}{n}$, for $r=1,2,\dots,n-1$:
$A=\sum_{r=1}^{n-1}\frac1n\left[\left(\frac{r}{n}\right)^{2}+3\left(\frac{r}{n}\right)\right]=\frac1n\left[\frac{1}{n^{2}}\sum_{r=1}^{n-1}r^{2}+\frac{3}{n}\sum_{r=1}^{n-1}r\right].$

2. Substitute the sum formulae

Using the given result (with upper limit $n-1$): $\sum_{r=1}^{n-1}r^{2}=\tfrac16(n-1)n(2n-1)$, and $\sum_{r=1}^{n-1}r=\tfrac12(n-1)n$:
$A=\frac1n\left[\frac{1}{n^{2}}\cdot\frac{(n-1)n(2n-1)}{6}+\frac{3}{n}\cdot\frac{(n-1)n}{2}\right]=\frac1n\left[\frac{(n-1)(2n-1)}{6n}+\frac{3(n-1)}{2}\right].$

3. Simplify to the required closed form

Factor out $(n-1)$ and combine over a common denominator $6n$:
$A=\frac{(n-1)}{n}\left[\frac{2n-1}{6n}+\frac{3}{2}\right]=\frac{(n-1)}{n}\cdot\frac{(2n-1)+9n}{6n}=\frac{(n-1)(11n-1)}{6n^{2}}.$
Hence $A=\dfrac{(n-1)(11n-1)}{6n^{2}}$ as required.

4. Deduce the exact integral by letting n → ∞

As $n\to\infty$ the rectangle width $\tfrac1n\to0$, so the total rectangle area $A$ tends to the exact area under the curve. Therefore
$\int_{0}^{1}\!\left(x^{2}+3x\right)dx=\lim_{n\to\infty}A=\lim_{n\to\infty}\frac{(n-1)(11n-1)}{6n^{2}}=\lim_{n\to\infty}\frac{11n^{2}-12n+1}{6n^{2}}=\lim_{n\to\infty}\left(\frac{11}{6}-\frac{2}{n}+\frac{1}{6n^{2}}\right)=\frac{11}{6}.$