Answer to: Approximate F(x) for x = 0, 0.5, 1, 1.5, 2, 2.5. F(x) = \int^x_0\sin(t^2)dt Round your answers to three decimal places. F(0) = ___ ...
10.求下列极限:(1)lim_(x→0)(∫_0^xsint^2dt)/(x^3) lim_(x→∞)x∫_0^x(t^2)/(√(1+t^2))dt ;sin x tan t dt(3)lim(4)lim x0+tan x x→0;√sin t dt sin t'dt arctan(1 +t) dt ] du(5)lim x(1- cos x)5sin(6)lim n+n+1+… + sin πn ...
这不叫 “定积分求导”,而是积分上限函数求导.记 F(x) = ∫[0,x]sin(t^2)dt,则 F'(x) = sin(x^2), 于是 (d/dx)∫[0,x^2]xsin(t^2)dt = (d/dx)x∫[0,x^2]sin(t^2)dt = (d/dx)[xF(x^2)] ... 分析总结。 过程尽量详细一点刚开始学还打算提炼方法谢了结果...
f(x)=int(0)^(x)(sint)/t dt के प्रांत में x gt0 के शीर्ष बिंदु है
If f(x) = int_0^xt sin t dt then find f'(x). View Solution ਜਦੋਂ ਕਿf(x)=∫x0tsintdt, ਤਦf'(x)ਹੈ: View Solution Iff(x)=∫x0{f(t)}−1dtand∫10{f(t)}−1=√2,then View Solution If f(t) is an odd function then∫x0f(t(dtis ...
syms x expr = x*log(1+x); F = int(expr,[0 1]) F = 14 Integrate another expression from sin(t) to 1. Get syms t F = int(2*x,[sin(t) 1]) F = cos(t)2 When int cannot compute the value of a definite integral, numerically approximate the integral by using vpa. Get ...
结果1 题目 2、微积分基本公式求$$ \frac { d } { d x } \int _ { 0 } ^ { x ^ { 2 } } \sin t ^ { 2 } d $$$ \frac{d}{1x}\int _{0}^{x^{2}}\sin t^{2}dt.-2=5^{3} $$ 相关知识点: 试题来源: 解析 微积分基本公式。从。 反馈 收藏...
百度试题 结果1 题目普通高等教育数学类基础课程$$ \frac{\int _{0}^{x2}\sin t^{2}dt}{x^{2}} $$ 相关知识点: 试题来源: 解析 高等数学(经管类下第3版普通高等教育数学基础课程十二五规划教材)/同济数学系列丛书。 反馈 收藏
Answer to: Find the derivative of F(x) if F(x) = int 0 4x sin(t) dt. a) -cos(4x) b) cos(4x) c) sin(4x) d) 4sin(4x) e) 4cos(4x) By signing up,...
वक्र y=int0^x।t। dt की एक स्पर्श रेखा दुवारा जो रेखा y=2x के समान्तर है, x- अक्ष पर काटा ग