Current Affairs JEE Main & Advanced

(1) \[\int_{0}^{\infty }{{{e}^{-ax}}\sin bxdx}=\frac{b}{{{a}^{2}}+{{b}^{2}}}\]           (2)  \[\int_{0}^{\infty }{{{e}^{-ax}}\cos bxdx}=\frac{a}{{{a}^{2}}+{{b}^{2}}}\]          (3)  \[\int_{0}^{\infty }{{{e}^{-ax}}}{{x}^{n}}dx=\frac{n!}{{{a}^{n}}+1}\]

\[\int_{0}^{\pi /2}{{{\sin }^{n}}xdx}=\int_{0}^{\pi /2}{{{\cos }^{n}}xdx}\]     \[\int_{0}^{\pi /2}{{{\sin }^{n}}xdx}=\int_{0}^{\pi /2}{{{\cos }^{n}}xdx}=\left\{ \begin{matrix} \frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4}......\frac{2}{3},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{when }n\text{ is odd}  \\ \frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4}.......\frac{3}{4}.\frac{1}{2}.\frac{\pi }{2},\,\,\,\,\,\text{when }n\text{ is even}  \\ \end{matrix} \right.\]     \[\int_{0}^{\pi /2}{{{\sin }^{m}}x{{\cos }^{n}}dx}=\frac{(m-1)\,(m-3).....(n-1)\,(n-3)....}{(m+n)\,(m+n-2)\,...(2\text{ or }1)}\],  [If \[m,\,\,n\] are both odd positive integers or one odd positive integer]     \[\int_{\,0}^{\,\pi /2}{{{\sin }^{m}}x{{\cos }^{n}}xdx}=\frac{(m-1)\,(m-3)............(n-1)\,(n-3)}{(m+n)\,(m+n-2)........(2\text{ or }1)}.\frac{\pi }{2}\], [If \[m,\,\,n\] are both positive integers]

(1) If  \[f(x)\] is continuous and \[u(x),\,\,v(x)\] are differentiable functions in the interval \[[a,\,\,b]\] then,     \[\frac{d}{dx}\int_{u(x)}^{v(x)}{f(t)dt=f\{v(x)\}\frac{d}{dx}}\{v(x)\}-f\{u(x)\}\frac{d}{dx}\{u(x)\}\].     (2) If the function \[\varphi \,(x)\] and \[\psi \,(x)\] are defined on \[[a,\,\,b]\] and differentiable at a point \[x\in \,(a,b),\] and \[f(x,t)\] is continuous, then, \[\frac{d}{dx}\,\left[ \int_{\varphi (x)}^{\psi (x)}{{}}f(x,t)\,dt \right]\]\[=\int_{\varphi (x)}^{\psi (x)}{\frac{d}{dx}}\,f(x,t)\,dt+\left\{ \frac{d\,\psi (x)}{dx} \right\}\,f(x,\psi (x))\]\[-\left\{ \frac{d\varphi (x)}{dx} \right\}f(x,\,\varphi (x))\].

(1) When both curves intersect at two points and their common area lies between these points: If the curves \[{{y}_{1}}={{f}_{1}}(x)\] and \[{{y}_{2}}={{f}_{2}}(x),\] where\[\,{{f}_{1}}(x)\,>\,{{f}_{2}}(x)\] intersect in two points \[A(x=a)\] and \[B(x=b)\], then common area between the curves is  \[=\int\limits_{a}^{b}{({{y}_{1}}-{{y}_{2}})\,dx}\]\[=\int\limits_{a}^{b}{[{{f}_{1}}(x)-{{f}_{2}}(x)]\,dx}\].     (2) When two curves intersect at a point and the area between them is bounded by x-axis: Area bounded by the curves \[{{y}_{{}}}={{f}_{1}}(x),{{y}_{2}}={{f}_{2}}(x)\,\,\text{and}\,x-\text{axis}\]is  \[\int\limits_{a}^{\alpha }{{{f}_{1}}(x)dx+\int\limits_{\alpha }^{b}{{{f}_{2}}(x)dx}}\],     where \[P(\alpha ,\beta )\,\]is the point of intersection of the two curves.     (3) Positive and negative area : Area is always taken as positive. If some part of the area lies above the x-axis and some part lies below x-axis, then the area of two parts should be calculated separately and then add their numerical values to get the desired area.

(1) If \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}xdx}\] then \[{{I}_{n}}+{{I}_{n-2}}=\frac{1}{n-1}\]     (2) If \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\cot }^{n}}xdx}\] then \[{{I}_{n}}+{{I}_{n-2}}=\frac{1}{1-n}\]     (3) If \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\sec }^{n}}x\,dx}\] then \[{{I}_{n}}=\frac{{{(\sqrt{2})}^{n-2}}}{n-1}+\frac{n-2}{n-1}{{I}_{n-2}}\]     (4) If \[{{I}_{n}}=\int_{0}^{\pi /4}{\text{cose}{{\text{c}}^{\text{n}}}x\,dx}\] then \[{{I}_{n}}=\frac{{{(\sqrt{2})}^{n-2}}}{n-1}+\frac{n-2}{n-1}{{I}_{n-2}}\]     (5) If \[{{I}_{n}}=\int_{0}^{\pi /2}{{{\sin }^{n}}x\,}dx,\,\] then \[{{I}_{n}}=\frac{n-1}{n}{{I}_{n-2}}\]     (6) If \[{{I}_{n}}=\int_{0}^{\pi /2}{{{\cos }^{n}}x\,dx,}\] then \[{{I}_{n}}=\frac{n-1}{n}{{I}_{n-2}}\].     (7) If \[{{I}_{n}}=\int_{0}^{\pi /2}{{{x}^{n}}\sin x\,dx}\] then \[{{I}_{n}}+n(n-1){{I}_{n-2}}=n{{(\pi /2)}^{n-1}}\]     (8) If \[{{I}_{n}}=\int_{0}^{\pi /2}{{{x}^{n}}\cos x\,dx}\] then \[{{I}_{n}}+n(n-1){{I}_{n-2}}=\,{{(\pi /2)}^{n}}\]     (9)  If \[a>b>0,\] then \[\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x}}=\frac{2}{\sqrt{{{a}^{2}}-{{b}^{2}}}}{{\tan }^{-1}}\sqrt{\frac{a+b}{a-b}}\]     (10)If \[n\,\in \,I\] then \[\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x}}=\frac{1}{\sqrt{{{b}^{2}}-{{a}^{2}}}}\log \left| \frac{\sqrt{b+a}-\sqrt{b-a}}{\sqrt{b+a}+\sqrt{b-a}} \right|\]     (11) If \[a>b>0\] then \[\int_{0}^{\pi /2}{\frac{dx}{a+b\sin x}=\frac{2}{\sqrt{{{a}^{2}}-{{b}^{2}}}}{{\tan }^{-1}}\sqrt{\frac{a-b}{a+b}}}\]     (12) If \[0<a<b\], then     \[\int_{0}^{\pi /2}{\frac{dx}{a+b\sin x}}=\frac{1}{\sqrt{{{b}^{2}}-{{a}^{2}}}}\log \left| \frac{\sqrt{b+a}+\sqrt{b-a}}{\sqrt{b+a}-\sqrt{b-a}} \right|\]     (13) If \[a>b,\,{{a}^{2}}>{{b}^{2}}+{{c}^{2}},\] then \[\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x+c\sin x}}\]\[\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x+c\sin x}}=\frac{2}{\sqrt{{{a}^{2}}-{{b}^{2}}-{{c}^{2}}}}\,{{\tan }^{-1}}\frac{a-b+c}{\sqrt{{{a}^{2}}-{{b}^{2}}-{{c}^{2}}}}\]     (14) If \[a>b,\,{{a}^{2}}<{{b}^{2}}+{{c}^{2}},\] then \[\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x+c\sin x}}\] \[=\frac{1}{\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}}\] \[\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x+c\sin x}=\frac{1}{\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}}}\log \left| \frac{a-b+c-\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}}{a-b+c+\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}} \right|\]     (15) If \[a<b,\] \[{{a}^{2}}<{{b}^{2}}+{{c}^{2}}\] then \[\int_{0}^{\pi /2}{\frac{dx}{a+b\cos x+c\sin x}}\]\[=\frac{-1}{\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}}\log \left| \frac{b-a-c-\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}}{b-a-c+\sqrt{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}} \right|\].

If a plane curve is revolved about some axis in the plane of the curve, then the body so generated is known as solid of revolution. The surface generated by the perimeter of the curve is known as surface of revolution and the volume generated by the area is called volume of revolution.     For example, a right angled triangle when revolved about one of its sides (forming the right angle) generates a right circular cones.     (1) Volumes of solids of revolution     (i) The volume of the solid generated by the revolution, about the x-axis, of the area bounded by the curve \[y=f(x),\]  the ordinates at \[x=a,\,x=b\] and the x-axis is equal to \[\pi \int_{a}^{b}{{{y}^{2\,}}\,dx}\].            (ii) The revolution of the area lying between the curve \[x=f(y)\] the y-axis and the lines \[y=a\] and \[y=b\] is given by (interchanging \[x\] and \[y\] in the above formulae) \[\int_{a}^{b}{\pi \,{{x}^{2\,}}}\,dy\].     (iii) If the equation of the generating curve be given by \[x={{f}_{1}}(t)\] and \[y={{f}_{2}}(t)\] and it is revolved about x-axis, then the formula corresponding to \[\int_{a}^{b}{\pi \,{{y}^{2\,}}\,dx}\] becomes \[\int_{{{t}_{1}}}^{{{t}_{2}}}{\pi {{\{{{f}_{2}}(t)\}}^{2}}\,d\,\{{{f}_{1}}(t)\}}\],     where \[{{f}_{1}}\] and \[{{f}_{2}}\] are the values of t corresponding to \[x=a\]  and  \[x=b\].     (2) Area of surfaces of revolution     (i) The curved surface of the solid generated by the revolution, about the x-axis, of the area bounded by the curve \[y=f(x)\], the ordinates at \[x=a,\,\,x=b\] and the x-axis is equal to \[2\pi \int_{x=a}^{x=b}{\,\,\,y\,ds}\].     (ii) If the arc of the curve \[y=f(x)\] revolves about y-axis, then the area of the surface of revolution (between proper limits) \[=2\pi \int_{{}}^{{}}{x\,ds,}\] where \[ds=\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}\,dx\].     (iii) If the equation of the curve is given in the parametric form \[x={{f}_{1}}(t)\] and \[y={{f}_{2}}(t)\], and the curve revolves about x-axis, then we get the area of the surface of revolution\[=2\pi \int_{t={{t}_{1}}}^{t={{t}_{2}}}{yds\,}\]     \[=2\pi \int_{t={{t}_{1}}}^{t={{t}_{2}}}{\,{{f}_{2}}(t)ds}\]\[=2\pi \int_{{{t}_{1}}}^{{{t}_{2}}}{{{f}_{2}}(t)\sqrt{\left\{ {{\left( \frac{dx}{dt} \right)}^{2}}+{{\left( \frac{dy}{dt} \right)}^{2}} \right\}}\,dt}\],     where \[{{t}_{1}}\] and \[{{t}_{2}}\] are the values of the parameter corresponding to \[x=a\] and \[x=b\].     (3) Volume and surface of the frustum of a cone : If \[{{r}_{1}},\,{{r}_{2}}\] be the radii of the circular ends and k is the distance between centres of circular ends and l be the slant height, then     (i) Volume of frustum of cone \[=\frac{\pi k}{3}(r_{1}^{2}+{{r}_{1}}{{r}_{2}}+r_{2}^{2})\]     (ii) Curved surface area of frustum of cone \[=\pi ({{r}_{1}}+{{r}_{2}})\,l\]     (iii) Whole surface area of frustum of cone     \[=\pi ({{r}_{1}}+{{r}_{2}})\,l\,+\pi \,r_{1}^{2}+\pi r_{2}^{2}\].   (4) Volume and surface of the frustum of a sphere : Let the thickness of the frustum of sphere is k and radii of the circular ends of the frustum are \[{{r}_{1}}\] and \[{{r}_{2}}\], then     (i) Volume of the frustum of sphere \[=\frac{\pi k}{6}(3r_{1}^{2}+3r_{2}^{2}+{{k}^{2}})\]     (ii) Curved surface area more...

Any function \[f(x)\] which is discontinuous at finite number of points in an interval \[[a,\,\,b]\] can be made continuous in sub-intervals by breaking the intervals into these subintervals. If \[f(x)\] is discontinuous at points \[{{x}_{1}},\,\,{{x}_{2}},\,\,{{x}_{3}}..........{{x}_{n}}\] in \[(a,\,\,b)\], then we can define subintervals \[(a,{{x}_{1}}),({{x}_{1}},{{x}_{2}}).............({{x}_{n-1}},\,\,{{x}_{n}}),\,({{x}_{n}},b)\] such that \[f(x)\] is continuous in each of these subintervals. Such functions are called piecewise continuous functions. For integration of piecewise continuous function, we integrate \[f(x)\] in these sub-intervals and finally add all the values.

(1) The area bounded by a cartesian curve\[y=f(x)\], x-axis and ordinates \[x=a\] and \[x=b\] is given by        Area \[=\int_{a}^{b}{y\,dx}=\int_{a}^{b}{f(x)dx}\]             (2) If the curve \[y=f(x)\] lies below x-axis, then the area bounded by the curve \[y=f(x),\] the x-axis and the ordinates \[x=a\] and \[x=b\] is negative. So, area is given by \[\left| \int_{a}^{b}{y\,dx} \right|\].     (3) The area bounded by a cartesian curve\[x=f(y),\,\] y-axis and abscissae \[y=c\] and \[y=d\] is given by,     Area \[=\int_{c}^{d}{x\,dy=\int_{c}^{d}{f(y)dy}}\]             (4) If the equation of a curve is in parametric form, let \[x=f(t),\,\,y=g(t)\] then the area \[=\int_{a}^{b}{y\,dx}=\int_{{{t}_{1}}}^{{{t}_{2}}}{g(t)\,f'(t)\,dt}\] , where \[{{t}_{1}}\] and \[{{t}_{2}}\] are the values of \[t\]  respectively  corresponding to the values of \[a\] and \[b\] of \[x\].  

If the curve is symmetrical about a co-ordinate axis (or a line or origin), then we find the area of one symmetrical portion and multiply it by the number of symmetrical portions to get the required area.  

Let the circle be \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] and line mirror \[lx+my+n=0\]. In this condition, radius of circle remains unchanged but centre changes. Let the centre of image circle be \[({{x}_{1}},\,\,{{y}_{1}})\]. Slope of \[{{C}_{1}}{{C}_{2}}\times \] (slope of \[lx+my+n=0)=-1\]            …..(i)     and mid point of \[{{C}_{1}}(-g,\,-f)\] and \[{{C}_{2}}({{x}_{1}},{{y}_{1}})\] lies on \[lx+my+n=0\]     i.e.,  \[l\,\left( \frac{{{x}_{1}}-g}{2} \right)\,+m\,\left( \frac{{{y}_{1}}-f}{2} \right)\,+\,n=0\]                   …..(ii)     Solving (i) and (ii), we get \[({{x}_{1}},\,\,{{y}_{1}})\]     \[\therefore \] Required image circle is \[{{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}\],  where \[r=\sqrt{({{g}^{2}}+{{f}^{2}}-c)}\]