Mathematics Test - 6

Main Concept :
Newton Leibniz Theorem(1)  If $f\left(x\right)$ is continuous and $\mathit{\text{u}}\left(x\right),$ $\mathit{v}\left(x\right),$ are differentiable functions in the interval [a, b], then,

$\frac{\text{d}}{\text{dx }}\left({\int }_{\mathit{\text{u}}\left(x\right)}^{\mathit{v}\left(x\right)}f\left(t\right)dt\right)=f\left\{\mathit{\text{v}}\left(x\right)\right\}\frac{\text{d}}{\text{dx }}\left\{v\left(x\right)\right\}-f\left\{u\left(x\right)\right\}\frac{\text{ d}}{\text{dx}}\left\{u\left(x\right)\right\}\text{.}$

(2)  If the function $\varphi \left(x\right) and \psi \left(x\right)$ are defined on [a, b] and differentiable at a point $xϵ\left(a,b\right),andf\left(x,t\right)$ is continuous, then,

$\frac{\text{d}}{\text{dx }}\left[{\int }_{\varphi \left(x\right)}^{\psi \left(x\right)}f\left(x,\text{t}\right) dt\right]= {\int }_{\varphi \left(x\right)}^{\psi \left(x\right)}\frac{\text{d}}{\text{dx }}f\left(x,t\right) dt+\left\{\frac{d\psi \left(\mathrm{ x}\right)}{\text{dx}}\right\}\mathrm{ f}\left(x,\psi \left(x\right)\right)-\left\{\frac{d \varphi \left(\mathrm{ x}\right)}{\text{dx}}\right\}\mathrm{ f}\left(x,\varphi \left(x\right)\right)\text{.}$

 

$\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}=\underset{0}{\overset{4}{\int }}\mathrm{x}\mathrm{d}\mathrm{y}=\underset{0}{\overset{4}{\int }}{\mathrm{y}}^2\mathrm{d}\mathrm{y}$

$={\left[\frac{y^3}{3}\right]}_0^4=\frac{64}{3}$
Main Concept :
Area with y axisThe area bounded by a cartesian curve x = f (y), y - axis and abscissae y = c and y = d is given by

$Area={\int }_c^{\text{d}}\left|x\right|dy={\int }_c^{\text{d}}\left|f\left(y\right)\right|dy$



Other Concepts :

Concept 1 :
Algebraic Integrals(1) Integral of the form $\int \frac{\mathit{d}\mathit{x}}{\sqrt{\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}}}$: To evaluate this form of integrals proceed as follows :

(i) Make the coefficient of unity by taking common from

Then, $\int \frac{dx}{\sqrt{a{x}^2+bx+c}}=\frac{1}{\sqrt{a}}\int \frac{dx}{\sqrt{x^2+\frac{b}{a} x+\frac{c}{a}}}$

(ii) Put by the method of completing the square in the form, $\sqrt{A^2-X^2} or \sqrt{X^2+A^2}$ or where, X is a linear function of x and A is a constant.

(iii) After this, use any of the following standard formulae according to the case under consideration

$\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\left(\frac{x}{a}\right)+c\Rightarrow \int \frac{dx}{\sqrt{x^2+a^2}}=\log \left|x+\sqrt{x^2+a^2} \right|+c$

$and \int \frac{dx}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+c.$

(2) Integral of the form $\int \sqrt{a{x}^2+bx+c} dx:$

This can be reduced to one of the forms of $\int \sqrt{a^2-x^2} dx, \int \sqrt{x^2-a^2} dx or \int \sqrt{a^2+x^2} dx.$

(3) Integral of the form $\int \frac{\left(px+q\right)dx}{a{x}^2+bx+c}, \int \frac{(px+q)dx}{\sqrt{a{x}^2+bx+c}} , \int \left(px+q\right)\sqrt{a{x}^2+bx+c} dx:$

For the evaluation of any of these integrals,

Put $px+q=A \left\{differentiation of \left(a{x}^2+bx+c\right)\right\}+B$

Find A and B by comparing the coefficients of like powers of x on the two sides.

In this way the integral breaks up into two parts.

(4)   Integrals of the form $\int \frac{\text{dx}}{\text{P}\sqrt{\text{Q}}}\text{.}$ (where P and Q are linear or quadratic expressions in x) : To evaluate such type of integrals, we have following substitutions according to the nature of expressions of P and Q in x :

(i)    When Q is linear and P is linear or quadratic, we put $\sqrt{Q }=t$.

(ii)   When P is linear and Q is quadratic, we put P = 1/t.

(iii)   When both P and Q are quadratic, we put x = 1/t

Let p(h, k) be the mid point of the line segment joining the focus (a, 0) and a general point Q (x,y) on the parabola. Then

Axis: The straight line passing through the focus and perpendicular to the directrix is called the axis of the conic section.

Vertex: The points of intersection of the conic section and the aixs is (are) called vertex (vertices) of the conic section.

Focal chord: Any chord passing through the focus is called focal chord of the conic section.

Double ordinate: A straight line drawn perpendicular to the axis and terminated at both end of the curve is a double ordinate of the conic section.

Latus rectum: The double ordinate passing through the focus is called the latus rectum of the conic section.

Center: The point which bisects every chord of the conic passing through it, is called the center of the conic section.

Other Concepts :

Concept 1 :
Parabola Symmetric about x-axis and its Parametric FormThe quadratic expression y² = 2px where P is the distance between focus and directrix, represents the source or the vertex form of the conic section called parabola with the vertex at the origin whose axis of symmetry coincide with the x - axis.

If we rewrite the above equation into x = ay² then,

Main Concept :
Functional EquationsIn mathematics, a functional equation is any equation that specifies a function in implicit form. Often, the equation relates the value of a function (or functions) at some point with its values at other points. For instance, properties of functions can be determined by considering the types of functional equations they satisfy. The term functional equation usually refers to equations that cannot be simply reduced to algebraic equations. An equation of the form f(x, y,.....) = 0, where f contains a finite number of independent variables, known functions, and unknown functions which are to be solved for. Many properties of functions can be determined by studying the types of functional equations they satisfy.
Other Concepts :

Concept 1 :
Examples on Functional Relations

Functional equations in more than one variable

For functional equations with more than one variable, we can also apply the methods mentioned above. Besides we can also try to substitute some special values, x = y = 0 into the given condition given to obtain some results. As it is very difficult to describe these techniques in words, we will try to see their application in various examples below.

Method of undermined coefficients

Sometimes after performing transformation of variables, we can arrive at simultaneous equaitons. We can find the unkonwn function after solving the simultaneous equations. We may also treat the unknown funtion as a variable in ordinary equations and solve it.

Remark,

In this question, we have a symmetric condition. By using the symmetry, we reduce the equation to a one - variable functional equation. This is a useful technique for symmetric functional equations.

Homogeneous form of Differential EquationsA differential equation of the form  are homogeneous series of  and  and of the same degree is called homogeneous differential equation is and can be solved by substituting y = vx

Other Concepts :

Concept 1 :
Solution of differential equations

Solution of differential equation in many type form of equation. 

(1) Differential equation of the first order and first degree.

(2) Equation in which the variables are separable.

(3) Differential equation reducible to the variable separable type.

(4) Homogeneous differential equation.

(5) First order liner differential equation.

(6) Exact differential equation.

(7) Bernoulli’s equation.