这意味着函数 f(x) = 1 在区间 [a, b] 上的定积分等于该区间的长度。例如，如果区间是 [0, 2]，则定积分 ∫_0^2 1 dx = 2 - 0 = 2。总结来说，定积分 ∫ 1 dx 的值为 x + C，其中 C 是积分常数。对于函数 f(x) = 1，其定积分在任意区间 [a, b] 上的值为 ...
View Solution ∫cosx1+cosxdx View Solution ∫cos−1xdx View Solution Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths Cengage Solutions for Maths DC Pandey Solutions for Physics HC Verma Solutions for Physics ...
int(dx)/((1+x^(4))^((1)/(4))) 09:04 int((x-1)/(x+1))^(4)dx 02:43 Evaluate each of the following integrals: intx^4\ dx (ii) intx^(5//... 02:46 int(dx)/((1-2^(x))^(4))dx 04:20 int(1)/(4^(-x)) dx 00:59 (i) int ((1+x)^(3))/(sqrt(x))dx...
{ 1 } { x ^ { \prime } } d x = \int _ { 0 } ^ { 1 + } \frac { 1 } { x } d x = \ln x | , = \lim i t s _ { x \rightarrow + \infty } \ln x - \ln 1 = + \infty $$,故积分 $$ f _ { 1 } ^ { 2 } $$ $$ \frac { 1 } ...
【题目】$$ ) \int \frac { x } { x ^ { 4 } + 1 } d x $$; 答案 【解析】 解:$$ \int \frac{x}{x^{4}+1}dx= \frac{1}{2}\int \frac{1}{x^{4}+1}dx^{2} $$ $$ = \frac{1}{2}\arctan x^{2}+C. $$相关...
Answer to: \int \frac{x dx}{1 + x^4} By signing up, you'll get thousands of step-by-step solutions to your homework questions. You can also ask...
∫1−1(x2+x+3)dx View Solution The value of the integral∫1−1(x−[2x])dx,is View Solution ∫1−1|2x+1|dx View Solution ∫1−1e2xdx View Solution The integral∫1−1|x+2|x+2dxis equal to View Solution Doubtnut is No.1 Study App and Learning App with Instant Video...
View Solution 1+cosec2xका न्यूनतम मान ज्ञात कीजिए। View Solution यदिy=1−cos2x1+cos2xहै तोdydxका मान ज्ञात कीजिए। ...
解(3)当 p1 时,(|sinxarctanx|)/(x^p)≤π/(2(x^y)) 而∫_1^(+∞)1/(x^p)dx 收敛,所以当 p1 时积分∫_1^(+∞)(sinxarcosx)/(x^p)dx 绝对收敛;当 0p≤1 时,因为积分∫_1^(+∞)(sinx)/(x^p)dx 收敛,arctanx在 [1,+∞) 单调有界,由Abel判别法,积分∫_1^(+∞)(sin...
【解析】 解:由定积分分部积分公式 $$ \int _ { 1 } ^ { e } x \ln x d x = \int _ { 1 } ^ { e } \ln x d \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 2 } } { 2 } \ln x | _ { 1 } ^ { e } - \int _ { 1 } ^ { e } \frac { x ^ { ...