Mathematics I (BT-102) Unit 2 Notes - RGPV

How unit 2 is examined

Integral calculus: the definite integral as a limit of a sum, Beta and Gamma functions, solids and surfaces of revolution, and double and triple integrals with change of order. Recent papers ask Beta-Gamma proofs, change of order and area or volume by integration.

Definite Integral as a limit of a sum and Its application in summation of series

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Definition. For a continuous $f$ on $[a,b]$, $\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{r=1}^{n} f(a+rh)\,h$ with $h=\frac{b-a}{n}$.

Key points.

The interval is cut into $n$ strips of width $h$, and the integral is the limit of the sum of strip areas as $n\to\infty$.
To sum a series, write the general term as $\frac1n f\!\left(\frac rn\right)$ and note that $r/n$ runs from $0$ to $1$.
Then $\lim_{n\to\infty}\frac1n\sum_{r=1}^{n} f\!\left(\frac rn\right)=\int_0^1 f(x)\,dx$, with $\frac rn\to x$, $\frac1n\to dx$ and $\sum\to\int$.
The limits are the lowest and highest values of $r/n$ (here $r/n$ for $r=1$ tends to $0$ and for $r=n$ equals $1$).

Example. $\lim_{n\to\infty}\left[\frac1{n+1}+\dots+\frac1{2n}\right]=\lim\frac1n\sum_{r=1}^{n}\frac{1}{1+r/n}=\int_0^1\frac{dx}{1+x}=\ln 2$.

Beta and Gamma functions and their properties

Definition. $\Gamma(n)=\int_0^\infty e^{-x}x^{n-1}dx$ $(n>0)$ and $B(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}dx$ $(m,n>0)$.

Key points.

$\Gamma(n+1)=n\,\Gamma(n)$, so $\Gamma(n+1)=n!$ for a positive integer $n$, and $\Gamma(1)=1$.
$\Gamma(\tfrac12)=\sqrt\pi$, and $\Gamma(n)\Gamma(1-n)=\dfrac{\pi}{\sin n\pi}$ for $0<n<1$.
$B(m,n)=B(n,m)$ and $B(m,n)=2\int_0^{\pi/2}\sin^{2m-1}\theta\cos^{2n-1}\theta\,d\theta$.
$B(m,n)=\int_0^\infty\frac{x^{m-1}}{(1+x)^{m+n}}dx$.
==$B(m,n)=\dfrac{\Gamma(m)\,\Gamma(n)}{\Gamma(m+n)}$.==

Proof of 5. Put $x=t^2$: $\Gamma(m)=2\int_0^\infty t^{2m-1}e^{-t^2}dt$. Then $\Gamma(m)\Gamma(n)=4\int_0^\infty\!\!\int_0^\infty x^{2m-1}y^{2n-1}e^{-(x^2+y^2)}dx\,dy$. Put $x=r\cos\theta,\ y=r\sin\theta$, $dx\,dy=r\,dr\,d\theta$: $$\Gamma(m)\Gamma(n)=\left[2\int_0^\infty r^{2m+2n-1}e^{-r^2}dr\right]\left[2\int_0^{\pi/2}\cos^{2m-1}\theta\sin^{2n-1}\theta\,d\theta\right]=\Gamma(m+n)\,B(m,n).$$

Proof of $\Gamma(n)\Gamma(1-n)$. Take $m=n$ and replace $n$ by $1-n$ in point 4, so $m+n=1$ and $\Gamma(1)=1$: $\Gamma(n)\Gamma(1-n)=B(n,1-n)=\int_0^\infty\frac{x^{n-1}}{1+x}dx=\frac{\pi}{\sin n\pi}$.

Example. $\int_0^1\sqrt{1-x^4}\,dx$: put $x^4=t$, $dx=\frac14 t^{-3/4}dt$, so the integral is $\frac14B(\tfrac14,\tfrac32)=\frac{\Gamma(1/4)\Gamma(3/2)}{4\,\Gamma(7/4)}$. With $\Gamma(\tfrac32)=\tfrac{\sqrt\pi}2$, $\Gamma(\tfrac74)=\tfrac34\Gamma(\tfrac34)$ and $\Gamma(\tfrac14)\Gamma(\tfrac34)=\pi\sqrt2$, this equals $\frac{\{\Gamma(1/4)\}^2}{6\sqrt{2\pi}}$.

Asked: [7 marks] (Nov 2022) Prove that $\Gamma(n)\Gamma(1-n)=\frac{\pi}{\sin n\pi}$ Asked: [7 marks] (Jun 2022) State and prove relationship between Beta and Gamma function. Asked: [7 marks] (Dec 2024) Prove that $\int_0^1\sqrt{1-x^4}\,dx=\frac{\{\Gamma(1/4)\}^2}{6\sqrt{2\pi}}$

Applications of definite integrals to evaluate surface areas and volumes of revolutions

Definition. Rotating a plane curve about a line sweeps out a solid of revolution; its volume and curved surface are found by integrating thin discs and thin bands.

Formula. About the x-axis: $V=\pi\int_a^b y^2dx$, $S=2\pi\int_a^b y\,ds$ with $ds=\sqrt{1+(dy/dx)^2}\,dx$. About the y-axis: $V=\pi\int x^2dy$, $S=2\pi\int x\,ds$. For a parametric curve $ds=\sqrt{x'^2+y'^2}\,dt$.

Key points.

A thin strip of the region becomes a disc of radius $y$ and thickness $dx$, of volume $\pi y^2dx$, and adding the discs gives $V$.
A small arc $ds$ becomes a band of area $2\pi y\,ds$, and adding the bands gives $S$.
Choose the axis of revolution first and integrate in the variable along that axis.
For a closed loop, revolve only the half on one side of the axis; revolving both halves counts the surface twice.

Example. Parabola $y^2=4ax$ cut by $x=a$, revolved about the tangent at the vertex (the y-axis): $x=\frac{y^2}{4a}$, so $V=\pi\int_{-2a}^{2a}\frac{y^4}{16a^2}dy=\frac{\pi}{16a^2}\cdot\frac{64a^5}{5}=$ $\frac{4\pi a^3}{5}$.

Example. Loop of $x=t^2,\ y=t-\frac{t^3}{3}$ about the x-axis: the loop is $0\le t\le\sqrt3$ (upper half), $x'^2+y'^2=4t^2+(1-t^2)^2=(1+t^2)^2$, so $ds=(1+t^2)dt$ and $S=2\pi\int_0^{\sqrt3}\left(t-\frac{t^3}3\right)(1+t^2)dt=2\pi\left[\frac{t^2}2+\frac{t^4}6-\frac{t^6}{18}\right]_0^{\sqrt3}=2\pi\cdot\frac32=$ $3\pi$.

Asked: [7 marks] (Dec 2024) The part of the parabola $y^2=4ax$ cut off by the latus rectum revolves about the tangent at the vertex. Find the volume of the solid generated. Asked: [7 marks] (Jun 2025) Show that the surface area of solid generated by revolution of the loop of curve $x=t^2,\ y=t-t^3/3$ about the x-axis is $3\pi$.

Multiple Integral

Definition. A double integral $\iint_R f(x,y)\,dA$ is the limit of the sum of $f\,\delta x\,\delta y$ over the region $R$, and a triple integral $\iiint_V f\,dV$ does the same over a volume.

Key points.

A double integral is evaluated as an iterated integral: $\int_a^b\int_{g_1(x)}^{g_2(x)} f\,dy\,dx$, integrating the inner variable first, treating the other as constant.
The inner limits may depend on the outer variable, but the outermost limits must be constants.
For a region bounded by constants, the order does not matter and the integral splits into a product when $f=g(x)h(y)$.
In polar form $x=r\cos\theta,\ y=r\sin\theta$ and $dA=r\,dr\,d\theta$.

Change the order of the integration

Definition. Changing the order means re-describing the same region $R$ with the other variable outside, so the integral becomes easier or possible.

Steps.

Step 1: From the given limits write the boundary curves of the region.
Step 2: Sketch the region and mark the intersection points.
Step 3: Draw a strip in the new direction and read the new limits (split the region if the strip changes its boundary).
Step 4: Integrate with the new limits.

Key points.

Points 1-3 of the steps fix the new limits, and the same region gives the same value.
Use it when the inner integral cannot be found, such as $\int\frac{dy}{\log y}$ or $\int e^{y^2}dy$.
If the new strip meets two different curves, split the region into two integrals.

Example. $\int_0^1\!\int_{x^2}^{2-x}xy\,dy\,dx$: the region is bounded by $y=x^2$ and $y=2-x$ meeting at $(1,1)$. Reversed: $\int_0^1\!\int_0^{\sqrt y}xy\,dx\,dy+\int_1^2\!\int_0^{2-y}xy\,dx\,dy=\frac16+\frac5{24}=$ $\frac38$.

Example. $\int_0^1\!\int_{e^x}^{e}\frac{dy\,dx}{\log y}$: region $0\le x\le1,\ e^x\le y\le e$ is $1\le y\le e,\ 0\le x\le\log y$, so $\int_1^e\frac{\log y}{\log y}dy=$ $e-1$.

Example. $\int_0^1\!\int_0^{\sqrt{1-x^2}}y^2dy\,dx$ is a quarter circle; reversed it is $\int_0^1\!\int_0^{\sqrt{1-y^2}}y^2dx\,dy=\int_0^1y^2\sqrt{1-y^2}\,dy=$ $\frac{\pi}{16}$.

Example. $\int_0^{4a}\!\int_{x^2/4a}^{2\sqrt{ax}}dy\,dx$: curves $y=\frac{x^2}{4a}$ and $y^2=4ax$ meet at $(0,0),(4a,4a)$. Reversed: $\int_0^{4a}\!\int_{y^2/4a}^{2\sqrt{ay}}dx\,dy=\int_0^{4a}\left(2\sqrt{ay}-\frac{y^2}{4a}\right)dy=$ $\frac{16a^2}{3}$.

Asked: [7 marks] (Jun 2025) Change the order of integration in $\int_0^1\int_{e^x}^e\frac{dy\,dx}{\log y}$ and then evaluate. Asked: [7 marks] (Nov 2022) By changing the order of integration, evaluate $\int_0^1\int_0^{\sqrt{1-x^2}}y^2\,dy\,dx$. Asked: [7 marks] (Jun 2022) Change the order of integration and evaluate $\int_0^{4a}\int_{x^2/4a}^{2\sqrt{ax}}dy\,dx$. Asked: [7 marks] (Dec 2023, Jun 2023) Change the order of integration in $\int_0^1\int_{x^2}^{2-x}xy\,dy\,dx$ and hence evaluate. Pitfall: Do not skip the sketch; wrong limits, or forgetting to split the region, is the usual loss of marks.

Applications of multiple integral for calculating area and volumes of the curves

Definition. Area of a plane region $R=\iint_R dx\,dy$ and volume under a surface $z=f(x,y)$ over $R$ is $\iint_R z\,dx\,dy$; a solid's volume is $\iiint dx\,dy\,dz$.

Key points.

In polar form area $=\iint r\,dr\,d\theta$.
For a volume bounded by a plane above the region $R$ in the xy-plane, integrate $z$ (the height) over $R$.
Use symmetry: compute one quadrant or octant and multiply.
Find the limits from the sketch of $R$ as in change of order.

Example. Area of the ellipse quadrant $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$: $\int_0^a\!\int_0^{b\sqrt{1-x^2/a^2}}dy\,dx=\frac ba\int_0^a\sqrt{a^2-x^2}\,dx=\frac ba\cdot\frac{\pi a^2}{4}=$ $\frac{\pi ab}{4}$.

Example. Tetrahedron under $\frac xa+\frac yb+\frac zc=1$: $V=\int_0^a\!\int_0^{b(1-x/a)}c\left(1-\frac xa-\frac yb\right)dy\,dx=\frac{cb}{2}\int_0^a\left(1-\frac xa\right)^2dx=$ $\frac{abc}{6}$.

Asked: [7 marks] (Nov 2022) Find the area of a plane in the form of a quadrant of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Asked: [7 marks] (Jun 2022) Using double integration, find the volume of the tetrahedron bounded by the coordinate planes and the plane $\frac xa+\frac yb+\frac zc=1$.

Last-minute revision

$\lim\frac1n\sum f(r/n)=\int_0^1f(x)\,dx$; replace $r/n$ by $x$ and $1/n$ by $dx$.
$\Gamma(n)=\int_0^\infty e^{-x}x^{n-1}dx$; $\Gamma(n+1)=n\Gamma(n)$; $\Gamma(\tfrac12)=\sqrt\pi$.
$B(m,n)=\dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}=2\int_0^{\pi/2}\sin^{2m-1}\theta\cos^{2n-1}\theta\,d\theta$.
$\Gamma(n)\Gamma(1-n)=\pi/\sin n\pi$.
$V_x=\pi\int y^2dx$, $S_x=2\pi\int y\,ds$; parametric $ds=\sqrt{x'^2+y'^2}\,dt$.
Parabola $y^2=4ax$ up to latus rectum about the y-axis: $V=\frac{4\pi a^3}5$.
Loop $x=t^2,y=t-t^3/3$ about the x-axis: $S=3\pi$.
Polar double integral: $dA=r\,dr\,d\theta$.
Change of order: sketch, redraw strip, split if two curves.
Ellipse quadrant $\frac{\pi ab}{4}$; tetrahedron $\frac{abc}6$.

Memory hooks

Gamma is the factorial with a shift: $\Gamma(n)=(n-1)!$.
Beta is a ratio of gammas: "top two over the sum".
Disc is $\pi y^2dx$, band is $2\pi y\,ds$.
Inner limits may move, outer limits stay fixed; if the inner integral is stuck, flip the order.
The tetrahedron has the "one-sixth" volume.

Coverage checklist

Definite Integral as a limit of a sum and Its application in summation of series: limit-of-sum formula and series example.
Beta and Gamma functions and their properties: Nov 2022, Jun 2022, Dec 2024 questions.
Applications of definite integrals to evaluate surface areas and volumes of revolutions: Dec 2024, Jun 2025 questions.
Multiple Integral: definition, iterated evaluation.
Change the order of the integration: Jun 2025, Nov 2022, Jun 2022, Dec 2023, Jun 2023 questions.
Applications of multiple integral for calculating area and volumes of the curves: Nov 2022, Jun 2022 questions.

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Mathematics I (BT-102) - Unit 2 Short Notes

How unit 2 is examined

Definite Integral as a limit of a sum and Its application in summation of series

Beta and Gamma functions and their properties

Applications of definite integrals to evaluate surface areas and volumes of revolutions

Multiple Integral

Change the order of the integration

Applications of multiple integral for calculating area and volumes of the curves

Last-minute revision

Memory hooks

Coverage checklist