7. (i) Evaluate $\iint_R e^{2x + 3y} \, dx \, dy$ over the triangular region bounded by x = 0, y = 0, and x + y = 1.

1. Set up the Integral:

   The region R is a right triangle with vertices at (0,0), (1,0),and (0,1). Therefore, x ranges from 0 to 1, and for a fixed x, y ranges from 0 to 1 - x.

2. Write the Limits of Integration:

   $\iint_R e^{2x + 3y} \, dx \, dy = \int_{0}^{1} \int_{0}^{1-x} e^{2x + 3y} \, dy \, dx$

3. Integrate with Respect to y:

   $\int_{0}^{1} \int_{0}^{1-x} e^{2x + 3y} \, dy \, dx$

   Compute the inner integral first:

   $\int_{0}^{1-x} e^{2x + 3y} \, dy$

   Since 2x is a constant with respect to y, we can treat it as a constant during integration:

   $\int_{0}^{1-x} e^{2x} e^{3y} \, dy = e^{2x} \int_{0}^{1-x} e^{3y} \, dy$

   The integral of $e^{3y}$ with respect to y is:

   $\int e^{3y} \, dy = \frac{e^{3y}}{3}$

   Evaluating from 0 to 1 - x:

   $e^{2x} \left[ \frac{e^{3y}}{3} \right]_{0}^{1-x} = e^{2x} \left( \frac{e^{3(1-x)}}{3} - \frac{e^{0}}{3} \right)$

   Simplify the expression:

   $e^{2x} \left( \frac{e^{3-3x}}{3} - \frac{1}{3} \right) = \frac{e^{2x} e^{3-3x}}{3} - \frac{e^{2x}}{3} = \frac{e^{3-x}}{3} - \frac{e^{2x}}{3}$

4. Integrate with Respect to x:

   Now we need to evaluate the outer integral:

   $\int_{0}^{1} \left( \frac{e^{3-x}}{3} - \frac{e^{2x}}{3} \right) dx$

   Split into two integrals:

   $\frac{1}{3} \int_{0}^{1} e^{3-x} \, dx - \frac{1}{3} \int_{0}^{1} e^{2x} \, dx$

   First Integral:

   $\frac{1}{3} \int_{0}^{1} e^{3-x} \, dx$

   Use substitution u = 3 - x, hence du = -dx:

   $\frac{1}{3} \int_{3}^{2} e^{u} (-du) = -\frac{1}{3} \int_{3}^{2} e^{u} \, du = -\frac{1}{3} \left[ e^{u} \right]_{3}^{2} = -\frac{1}{3} (e^{2} - e^{3}) = \frac{e^{3} - e^{2}}{3}$

   Second Integral:

   $-\frac{1}{3} \int_{0}^{1} e^{2x} \, dx$

   The integral of $e^{2x}$ with respect to x is:

   $\int e^{2x} \, dx = \frac{e^{2x}}{2}$

   Evaluating from 0 to 1:

   $-\frac{1}{3} \left[ \frac{e^{2}}{2} - \frac{1}{2} \right] = -\frac{1}{3} \cdot \frac{e^{2} - 1}{2} = -\frac{e^{2} - 1}{6}$

5. Combine the Results:

   Adding the results of the two integrals:

   $\frac{e^{3} - e^{2}}{3} - \frac{e^{2} - 1}{6} = \frac{2(e^{3} - e^{2}) - (e^{2} - 1)}{6} = \frac{2e^{3} - 2e^{2} - e^{2} + 1}{6} = \frac{2e^{3} - 3e^{2} + 1}{6}$

   Hence, the value of the integral is:

   $\boxed{\frac{2e^{3} - 3e^{2} + 1}{6}}$

(ii) Change the order of integration in the following integral and evaluate: $\int_{0}^{4a} \int_{\frac{x^2}{4a}}^{2\sqrt{xa}} dy \, dx$

Here integration is first w.r.t. $y$ and $P$ on the parabola $x^2 = 4ay$ to Q on the parabola $y^2 = 4ax$ and then w.r.t. x from $x = 0$ to $x = 4a$ giving the shaded region of integration (Fig.).

On changing the order of integration, we first integrate w.r.t. x from R to S , then w.r.t. y from y = 0 to y = 4a .

$\therefore \, I = \int_0^{4a} \left(\int_{y^2/4a}^{2\sqrt{ay}} dx \right) dy = \int_0^{4a} \left( 2\sqrt{ay} - \frac{y^2}{4a} \right) dy$

$= \left[ 2\sqrt{a} \cdot \frac{y^{3/2}}{3/2} - \frac{y^3}{12a} \right]_0^{4a}$

$= \left[\frac{32a^2}{3} - \frac{16a^2}{3} \right] = \frac{16a^2}{3}.$

8. (i) Evaluate $\int_0^2 \int_0^{\sqrt{2x - x^3}} \frac{x \, dy \, dx}{\sqrt{x^2 + y^2}}$ by changing to polar coordinates.

1. Convert the integrand to polar coordinates:

   $x = r \cos \theta$ and $y = r \sin \theta$

   then,

   $x^2 + y^2 = r^2$

   and the differential area element in polar coordinates is:

   $dx \, dy = r \, dr \, d\theta$

   So, the integrand $\frac{x \, dy \, dx}{\sqrt{x^2 + y^2}} = \frac{r \cos \theta \cdot r \, dr \, d\theta}{r} = r \cos \theta \, dr \, d\theta$ in polar coordinates becomes:

2. Determine the limits of integration:

   • For x from 0 to 2 and y from 0 to $\sqrt{2x - x^2}$

   • r goes from 0 to 2.

   • θ goes from 0 to $π/2$ (since the region is in the first quadrant).

3. Set up and evaluate the integral in polar coordinates:

   $\int_{0}^{2} \int_{0}^{\frac{\pi}{2}} r \cos \theta \, r \, d\theta \, dr$

4. Evaluation:

   First, evaluate the inner integral with respect to θ:

   $\int_{0}^{\frac{\pi}{2}} r \cos \theta \, d\theta$ $r \left[ \sin \theta \right]_{0}^{\frac{\pi}{2}}$ $r \left(\sin \frac{\pi}{2} - \sin 0 \right)$ $r (1 - 0) = r$ evaluate the outer integral with respect to r:

   $\int_{0}^{2} r \cdot r \, dr$ $\int_{0}^{2} r^2 \, dr$ $\left[ \frac{r^3}{3} \right]_{0}^{2}$ $\frac{2^3}{3} - \frac{0^3}{3}$ $\frac{8}{3}$

   Therefore, the value of the integral is: $\boxed{ \frac{8}{3}}$

(ii) Find the volume bounded by the cylinder $x^2 + y^2 = 4$ and the planes $y + z = 4$ and z = 0.

From Fig. , it is self-evident that z = 4 - y is to be integrated over the circle $x^2 + y^2 = 4$ in the XY-plane. To cover the shaded half of this circle, x varies from 0 to $\sqrt{(4 - y^2)}$ and y varies from -2 to 2.

$\therefore \, \text{Required volume}$

$= 2 \int_{-2}^{2} \left( \int_{0}^{\sqrt{(4 - y^2)}} z \, dx \, dy \right) = 2 \int_{-2}^{2} \left( \int_{0}^{\sqrt{(4 - y^2)}} (4 - y) \, dx \, dy \right)$

$= 2 \int_{-2}^{2} (4 - y) \left[ x \right]_0^{\sqrt{(4 - y^2)}} dy = 2 \int_{-2}^{2} (4 - y) \sqrt{(4 - y^2)} \, dy$

$= 2 \int_{-2}^{2} 4 \sqrt{(4 - y^2)} \, dy - 2 \int_{-2}^{2} y \sqrt{(4 - y^2)} \, dy$

$= 8 \int_{-2}^{2} \sqrt{(4 - y^2)} \, dy$

[The second term vanishes as the integrand is an odd function.]

$= 8 \left[ \frac{y \sqrt{(4 - y^2)}}{2} + \frac{4}{2} \sin^{-1} \frac{y}{2} \right]_{-2}^{2} = 16 \pi$

9. (i) Define Series with example.

A series is the sum of the terms of a sequence. Given a sequence $a_1, a_2, a_3, \ldots$, the corresponding series is written as:

$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots$

A series can be finite or infinite. If the series continues indefinitely, it is called an infinite series.

Example: Geometric Series

A common example of a series is the geometric series. A geometric series is one where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio r.

The general form of a geometric series is:

$\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \cdots$

where:

• a is the first term,
• r is the common ratio.

Convergence of a Geometric Series

The geometric series converges if $|r| < 1$ and diverges if $|r| \geq 1$.

If $|r| < 1$, the sum of the infinite geometric series is given by:

$\sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r}$

(ii) Define Cauchy integral test.

The Cauchy integral test is a criterion used to determine the convergence of certain infinite series. Specifically, it is useful for series whose terms are given by the values of a continuous, positive, and decreasing function.

Cauchy Integral Test

Let $f(x)$ be a continuous, positive, and decreasing function for $x \geq N$, where N is a positive integer. Consider the infinite series $\sum_{n=N}^{\infty} f(n)$. The convergence of this series can be determined by the convergence of the improper integral $\int_{N}^{\infty} f(x) \, dx$.

1. If $\int_{N}^{\infty} f(x) \, dx$ converges, then the series $\sum_{n=N}^{\infty} f(n)$ also converges.
2. If $\int_{N}^{\infty} f(x) \, dx$ diverges, then the series $\sum_{n=N}^{\infty} f(n)$ also diverges.

(iii) State Lagrange's Cauchy mean value theorem.

Lagrange's Mean Value Theorem (often simply called the Mean Value Theorem) is a fundamental result in calculus that provides a formalized way of understanding how a function behaves on an interval. It is named after Joseph-Louis Lagrange, who formulated this theorem in the context of differential calculus.

Lagrange's Mean Value Theorem

Let f be a function that satisfies the following conditions:

1. f is continuous on the closed interval $[a, b]$.
2. f is differentiable on the open interval $(a, b)$.

Then there exists at least one point c in $(a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

In other words, there exists a point c in $(a, b)$ where the instantaneous rate of change (the derivative) of the function is equal to the average rate of change of the function over the interval $[a, b]$.

(iv) Define continuity of a function with example.

A function f is said to be continuous at a point c in its domain if the following three conditions are satisfied:

1. Existence of $f(c)$: The function f is defined at c; that is, f(c) exists.

2. Existence of the Limit: The limit of f(x) as x approaches c exists.

$\lim_{x \to c} f(x) \text{ exists}$

3. Limit Equals the Function Value: The limit of f(x) as x approaches c is equal to the function value at c.

$\lim_{x \to c} f(x) = f(c)$

If f is continuous at every point in its domain, then f is said to be a continuous function.

(v) Define Taylor Series.

A Taylor series is an infinite series representation of a function that is expressed as a sum of terms calculated from the values of the function's derivatives at a single point. For a function f(x), its Taylor series expansion around a point a is given by:

$f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots$ where f(a) is the value of the function at x=a, f′(a) is the first derivative of f evaluated at x=a, f′′(a) is the second derivative evaluated at x=a, and so on. This series allows one to approximate the function f(x) near the point a using a polynomial whose terms are determined by the derivatives of f at that point.

(vi) Define absolute convergence.

Absolute convergence is a property that applies to series in mathematics, particularly infinite series. A series $\sum a_n$ is said to converge absolutely if the series formed by taking the absolute values of its terms, $\sum |a_n|$ converges. In other words, a series converges absolutely if the sum of the absolute values of its terms converges.

Mathematically,$\sum a_n$ converges absolutely if $\sum |a_n|$ converges:

$\sum_{n=1}^{\infty} a_n \text{ converges absolutely} \iff \sum_{n=1}^{\infty} |a_n| \text{ converges}.$

Absolute convergence is a stronger condition than regular convergence because every absolutely convergent series is also convergent, but the converse is not necessarily true.

(vii) Define double integral.

A double integral is an extension of the concept of a single integral to functions of two variables over a region in the plane. For a function $f(x,y)$, the double integral over a region D in the xy-plane is denoted as:

$\iint_D f(x, y) \, dx \, dy$

• D represents the region of integration in the xy-plane.

• f(x,y) is the integrand, which specifies the function being integrated over the region D.

• dxdy indicates the order of integration, with dx representing integration with respect to x first, followed by integration with respect to y.

  The double integral computes the volume under the surface z=f(x,y) over the region D in the xy-plane. It can also be evaluated by reversing the order of integration (integrating with respect to y first, then x) if the region D is more naturally described in terms of y-bounds varying with x.

(viii) Define alternating Series.

An alternating series is a series where the terms alternate in sign. This means that consecutive terms switch between positive and negative. A general form of an alternating series can be written as:

$\sum_{n=1}^{\infty} (-1)^{n-1} a_n$

or,

$\sum_{n=1}^{\infty} (-1)^{n} a_n$

where $a_n$ is a sequence of positive real numbers.

(ix) Define definite integral

A definite integral is a fundamental concept in calculus that represents the accumulation of a quantity, which can be interpreted geometrically as the area under a curve. Formally, the definite integral of a function f(x) over an interval [a,b] is denoted by:

$\int_{a}^{b} f(x) \, dx$

This integral gives the total accumulation of f(x) from x=a to x=b. The limits a and b are called the lower and upper limits of integration, respectively.

(x) Define change of order of integration in double integral.

The change of order of integration in a double integral refers to switching the sequence in which the integrations are performed. This technique is useful when evaluating a double integral where the original order of integration is difficult or impossible to compute directly. By changing the order of integration, we often simplify the integral or make it possible to solve.

Consider a double integral of a function f(x, y) over a region R in the xy-plane. The double integral is expressed as:

$\iint_{R} f(x, y) \, dA$

If the region R is described as:

1. Type I Region: R can be represented as:

   $R = \{(x, y) \mid a \leq x \leq b, \, g_1(x) \leq y \leq g_2(x)\}$

   The double integral is written as:

   $\int_{a}^{b} \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \, dx$

2. Type II Region: R can be represented as: $R = \{(x, y) \mid c \leq y \leq d, \, h_1(y) \leq x \leq h_2(y)\}$ The double integral is written as: $\int_{c}^{d} \int_{h_1(y)}^{h_2(y)} f(x, y) \, dx \, dy$