int-4^4[tan^-1(frac{x^2}{x^4+1})+tan^-1(frac{x^4+1}{x^2})]dx இன் மதிப்... 03:47 int-frac{pi}{4}^frac{pi}{4}(frac{2x^7-3x^5+7x^3-x+1}{cos^2x))dx இன் மத... 05:13 f(x)=int0^xtcostdt, எனில் frac{df}...
Step by step video & image solution for Let f(x)=intx^(sinx)(1+xcosxdotlnx+sinx)dxa n df(pi/2)=(pi^2)/4dot Then the value of |"cos"(f(pi))| is___ by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams.Updated on:21/07/2023Class...
设f(x)在 [a,b] 上连续,且严格单调增加,证明: (a+b)∫_a^bf(x)dx2∫_a^bxf(x)dx . 答案 证明见分析证作辅助函数 F(x)=(a+x)∫_a^xf(t)dt-2∫_a^xtf(t) dt,则 f'(x)=∫_a^xf(t)dt+(a+x)f(x)-2xf(x)=∫_a^xf(t)dt+(a-x)f(x)df(x)=f(x)f(x)=f...
证明:(1)取M=max{|a|,1},则由xf(x)dx收敛,可知xf(x)dx也收敛,而0≤Mf()x≤xf()dx,故f()dx收敛,从而f)dx收敛(2)由于在[a,+∞)上f(x)与f(x)均为连续函数,任给Aa,有SAaf'(dx-jardf()-xf-/()fx)dx收敛,由f(x)的单调性可知,limxf (x)=-af(a).从而可知设lmxf(x)...
Answer to: Differentiate w respect to x. \frac{df}{dx}f(x) = \int_a^{x^2} \sqrt{t^e - 2t} dt Don't forget to check continuity. By signing up,...
Find f'(x). f(x) = ln^2(x). Find df/dx and df/dy when f(x,y) = (4x^4 y^3 + 10)^2. Find f'(x): f(x)=x^e+ \frac{1}{x^{\sqrt {10} If f(x) = (x^2 + 2x + 8)^3, find f'(x) and f'(5). Find f'(x) if f(x) = 11/x^10. Find f'(x). ...
{eq}\displaystyle \iint\limits_Sf(x, y, z)\,d\mathbf{S} = \iint\limits_Df(x, g(x, z), z)\sqrt{\left(\frac{\partial y}{\partial x}\right)^2+ 1 + \left(\frac{\partial y}{\partial z}\right)^2}\, dA {/eq}. Ans...
∴$$ f(x)= \frac{1- \ln x}{x^{2}}+c $$ $$ \intxf^{\prime}(x)dx= \int \times df(x)=xf(x)- \int f(x) \\ 分解积分法 = \frac{1- \ln x}{x}- \frac{\ln x}{x}+C \\ = \frac{1-2 \ln x}{x}+c $$ ...
(x) ,则∫f^(-1)(x)dx=xf^(-1)(x)-F(f^(-1)(x))+C 方法二:由分部积分法可得因为f1(x)是f(x)的反函数,且∫f(x)dx=F(x)+C ,所以x=f(f1(x)=F′(f-1(x),于是∫f^(-1)(x)dx=xf^(-1)(x)-∫(F'(f^(-1))(x)df^(-1)(x)) .=xf^(-1)(x)-F(f^(...
Answer to: If f(x) = \displaystyle \int_{\ln x}^{e^2} \dfrac{\cos t}{t}\,dt, then \dfrac{df}{dx} = ? By signing up, you'll get thousands of...