Recently Active Calculus Questions

*Image* for PRACTICE ...

\int_{-\pi /2}^{\pi /2}{\frac{cosx}{1+e^{x}}}dx for practice.please post the solution too. ans------->1 ...

1) f'(x) = f(x) - ln x + 1/x 2) ( 1 + x√(x2 + y2))dx + ( -1 + √(x2 + y2) )y dy = 0 3) 2x3y dy +( 1 - y2)(x2y2 + y2 -1) dx = 0 [find the soln. of the above differential eqn.] [ PLEAZ DON'T ASK ME ANS. BECOZ I DON'T HAVE TH ...

i ll give a sum and post its solution!! explain to me y is it the way it is!! *Image* *Image* *Image* questions...can all summations be treated like this?? why for Tn n->0 but Sn n->∞ please help urgent IITJEE 2008 QU ...

\int_{0}^{3\alpha }{cosec(x-\alpha )cosec(x-2\alpha )}dx ans-------> 2cosec\alpha log(1/2 sec\alpha ) ...

if n>1 evaluate : \int_{0}^{\infty }{\frac{dx}{(x+\sqrt{1+x^{2}})^{n}}} ans -----> n/n2 - 1 ...

find all values of " a " for which the inequality \frac{1}{\sqrt{a}}\int_{1}^{a}\left( {\frac{3\sqrt{x}}{2}+1-\frac{1}{\sqrt{x}}\right)}<4 is satisfied. ans-------->(0,4) ...

Here goes my first question in TargetIIT :):):):):):) Find the following limits - 1 > lim [ ( n ) !/n n ] 2 > lim [ ( n - 1 ) !/n n - 1.5 ] e n In both cases n tends to infinity . ...

1. \int \frac{x^2}{\sqrt{1- x^2}} 2. \int \frac{x^3}{\sqrt{1- x^2}} 3 . \int \frac{dx}{( x^2 + 1 ) ^2 } = \frac{A}{148}\tan^{-1} x +\frac{x}{2(x^2 + 1)} find value of A 4 . \int \frac{(1+ logx )dx}{\sqrt{x^{2x}-1}} for practi ...

f(x)=sin x + x ; find ∫0pi f-1(x) dx. ...

Can anyone integrate , ln ( u2 - 2 u cos x + 1 ) dx ? Limits are from x = 0 to x = pi . ...

The area bounded by the curves y = sin-1 |sin x| and y = (sin-1 |sin x|)2 , x \epsilon [0 , 2 \pi ] is: a) 1/3 + \pi 2/4 b) 1/6 + \pi 2/8 c) 2/3 + \pi 2/2 d) none of these ----------------------------------------------------- ...

Dunno where i'm doing wrong... \lim_{x\rightarrow \propto } ln\left(x+\frac{1}{x} \right) ?? ...

x^{2}\left(\int_{0}^{\pi /2}{(2sint+3cost)dt} \right)-x\left(\int_{-3}^{3}{\frac{t^{2}sin2t}{t^{2}+1}}dt \right)-2=0, then x= ...... ans-----> \pm \sqrt{2/5} ...

\int_{0}^{\pi /4}{\left\{2sinxcosx \right\}}d\left(x-\left[x \right] \right) where [x] and{x} r G.I.F and fractional part respectively. ans-----> 1/2 ...

The least value of \phi (x) = \int_{x}^{2}{log_{1/3}tdt}for x\epsilon \left(\frac{1}{10},4 \right) is at x = ........ ans--->1 ...

\int_{-\pi /2}^{\pi /2}{\sqrt{cos^{2n-1}x-cos^{2n+1}x}}dx,n\epsilon N- ans------> 4/2n+1 ...

\int \left(\frac{1-\sqrt{x}}{1+\sqrt{x}} \right)^{1/2}\frac{dx}{x} just for practise ans--> -2log\frac{1+\sqrt{1-x}}{\sqrt{x}}-2sin^{-1}\sqrt{x}+c ...

∫log(1+x)/1+x2dx limits r 0 to pi/4 ...

P(x,y) is a point which moves in the x-y plane such that 2[y] = 3[x] , where [.] denotes GIF & -2<= x <=5 ; -3 <= y <= 6. The area of the region containing the point P(x,y) is equal to: a)2 b)6 c) 4 d)8 ---------- ...

here goes my doubt i read it in arihant according to complex numbers *Image* *Image* *Image* now consider *Image* *Image* please explain last two steps ...

1. I=\int cosh^2(8x+5)dx 2. I=\int \frac{dx}{(x^2+x+1)} 3. I=\int\frac{(2+3x^2)dx}{[x^2(1+x^2)]} .. 4. I=\int\frac{(x^2-1)dx}{(x^4+3x^2+1)arctan[(x^2+1)/x]} 5. I=\int \frac{(\sqrt{a^2-x^2})dx}{x^4} plzz help me m a begginer n ...

1) If f(x) = x/1+((log x)(log x)(log x)(log x)..... \propto ). for all values of x \epsilon [1 , \propto ) ; then \int_{1}^{2e}{f(x)dx} is equal to : a) e2 - 1/2 b) e2 + 1/2 c) e2 - 2e/2 d) none of these --------------------- ...

relaxing, i came about this term!! GAMMA FUNCTION!! can anyone explain its meaning??? ...

∫x2/(xsinx+cosx)2dx ...

∫√sin(x-a)/sin(x+a)dx ...

show that \int_{0}^{\pi /2}{sin\theta logsin\theta }d\theta = log\frac{2}{e} ...

Discuss the continuity of the function f(x)= x3+x2+1 ...

IF ∫(sin 3θ + sin θ)esin θ cos θ dθ = (Asin 3θ + Bcos 2θ + Csin θ + D cos θ + E)esin θ + c find A, B , C , D , E ANS---> -4 , -12 , -20 , 0 , 32 respectively ...

have you read reflections .... the magazine brought out by the admins of this site?? check it out.... http://reflections.targetiit.com/ well going through it i came across a sum and the solution... iam posting both of them... ...