Let `f` be a continuous function on `[a ,b]dot` Prove that there exists a number `x in [a , b]` such that `int_a^xf(t)dx=int_x^bf(t)dtdot`
1 Answers
Let `g(x)=int_(a)^(x)f(t)dt-int_(x)^(b)f(t)dt, xepsilon[a,b]`
We have `g(a)=int_(a)^(b)f(t)dt` and `g(b)=int_(a)^(b)f(t)dt`
`:.g(a).g(b)=-[int_(a)^(b)f(t)dt]^(2)le0`
clearly `g(x)` is continuous in `[a,b]` an `g(a).g(b)le0`
It implies that `g(x)` will become zero at least once in `[a,b]`.
Hence `int_(a)^(x)f(t)dt=int_(x)^(b)f(t)dt` for all least one value of `xepsilon[a,b]`
0
like
0 dislike
Talk Doctor Online in Bissoy App
Show that `int_a^b(|x|)/x dx=|b|=|a|dot`
Case I: If `0lealtb`, then `|x|//x=1` `:.I=int_(a)^(b)1dx=b-a=|b|-|a|` Case II: If `altble0`, then `|x|=-x` `:. I=int_(a)^(b)(-x)/xdx=int_(a)^(b)(-1)dx` `=[-x]_(a)^(b)=b-(-a)=|b|-|a|` Case III: If `alt0ltb` then `|x|=-x` when `altxlt0` and `|x|=x`, when `altxltb` `:.I=int_(a)^(|x|)/xdx=int_(a)^(0)(|x|)/x dx+int_(0)^(b)(|x|)/x...
2 Answers 1 views
Let `f` be a real-valued function satisfying `f(x)+f(x+4)=f(x+2)+f(x+6)` Prove that `int_x^(x+8)f(t)dt` is constant function.
Given that `f(x)+f(x+4)=f(x+2)+f(x+6)`…………1 Replacing `x` by `x+2` we get `f(x+2)+f(x+6)=f(x+4)+f(x+8)`…………………2 From equation 1 and 2 we get `f(x)=f(x+8)`……………3 or `int_(x)^(x+8)f(t)dt+int_(0)^(8)f(t)dt` Thus, `g` is a constant function.
2 Answers 1 views
For `x >0,l e tf(x)=int_1^x(logt)/(1+t)dtdot` Find the function `f(x)+f(1/x)` and find the value of `f(e)+f(1/e)dot`
Let `F(x)=(x)+f(1/x)` `=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(1//x)(logt)/(1+t)dt` In second integral let `t=1//y`. So `dt=-1/(y^(2))dy` `:.F(x)=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(x)(-logy)/(1+1/y)(-(dy)/(y^(2)))` `=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(x)(logy)/(y(1+y))dy` `=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(x)(logt)/(t(1+t))dt` `=int_(1)^(x)(logt)/t dt=1/2(logx)^(2)` `:.F(e)=1/2`
2 Answers 1 views
Find the value of `int_(-1)^1[x^2+{x}]dx ,w h e r e[dot]a n d{dot}` denote the greatest function and fractional parts of `x ,` respectively.
Correct Answer - `(3-sqrt(5))/2` `int_(-1)^(1)[x^(2)+{x}]dx=int_(-1)^(0)[x^(2)+x+1]dx+int_(0)^(1)[x^(2)+x]dx` `=0+int_(0)^((sqrt(5)-1)/2)0dx+int_((sqrt(5)-1)/2)^(1)1dx` `=(3-sqrt(5))/2`
2 Answers 1 views
`f(x)` is continuous function for all real values of `x` and satisfies `int_0^xf(t)dt=int_x^1t^2f(t)dt+(x^(16))/8+(x^6)/3+adot` Then the value of `a`
Correct Answer - D `int_(0)^(x)f(t)dt=int_(x)^(1)t^(2)f(t)dt+(x^(16))/8+(x^(6))/3+a`……………..1 For `x=1, int_(0)^(1)f(t)dt=01/8+1/3+a=11/24+a`……………2 Differentiating both sides of w.r.t`x` We get `f(x)=0-x^(2)f(x)+2x^(15)+2x^(5)` `impliesf(x)=(2(x^(15)+x^(4)))/(1+x^(2))` `implies2int_(0)^(1)(x^(15)+x^(5))/(1+x^(2))dx+11/24+a`(from a) `implies2int_(0)^(1)(x^(13)-x^(11)+x^(9)-x^(7)+x^(5))dx=11/24+a` `implies 2(1/14-1/12+1/10-1/8+1/6)=11/24+a` `implies a=-167/840`
2 Answers 1 views
Given that `f` satisfies `|f(u)-f(v)|lt=|u-v|forua n dv` in `[a , b]dot` Then `|int_a^bf(x)dx-b(b-a)f(a)|lt=` `((b-a))/2` (b) `((b-a)^2)/2` `(b-a)^2`
Correct Answer - B `|int_(a)^(b)f(x)dx-(b-a)f(a)|=|int_(a)^(b)f(x)dx-int_(1)^(b)f(a)dx|` `=|int_(a)^(b)(f(x)-f(a))dx|` `le int_(a)^(b)|f(x)-f(a)|dx` `le int_(a)^(b)|x-a|dx` `=int_(a)^(b)(x-a)dx=((b-a)^(2))/2`
2 Answers 1 views
`L e tI_1=int_(-2)^2(x^6+3x^5+7x^4)/(x^4+2)dxa n d` `I_2=int_(-3)^1(2(x+1)^2+11(x+1)+14)/((x+1)^4+2)dtdot` Then the value of `I_1+I_2i s` 8 (b) `(200)
Correct Answer - C In `I_(2)` put `x+1=t`. Then `I_(2)=int_(-2)^(2)(2t^(2)+11t+14)/(t^(4)+2)dt=int_(-2)^(2)(2x^(2)+11x+14)/(x^(4)+2)dx` `:. I_(1)+I_(2)=int_(-1)^(2)(x^(6)+3x^(5)+7x^(4)+2x^(2)+11x+14)/(x^(4)+2) dx` `int_(-2)^(2)((x^(2)+3x+7)(x^(4)+2)+5x)/(x^(4)+2)dx` `=int_(-2)^(2)(x^(2)+3x+7)dx+5int_(-2)^(2)x/(x^(4)+2)dx` `=2 int_(0)^(2)(x^(2)+7)dx=100/3`
2 Answers 1 views
If `int_a^b|sinx|dx=8` and `int_0^(a+b)|cosx| dx=9` , then find the value of `int_a^b xsinx dx`.
Correct Answer - A::B We know `int_(a)^(b)|sinx|dx` represents the area under the curve from `x=a` to `x=b`.We also know that area from `x=a` to `x=a+pi` is 2. `:. int_(a)^(b)|sinx|dx=8` or `b-a=(8pi)/2`……………1...
2 Answers 1 views
Let `f(x)` and `phi(x)` are two continuous function on `R` satisfying `phi(x)=int_(a)^(x)f(t)dt, a!=0` and another continuous function `g(x)` satisfyi
Correct Answer - B `f(x)` is an odd function. Thus, `f(x)=-f(-x)` `phi(-x)=int_(0)^(-x)f(t)dt` Put `t=-y` `:. phi(-x)=int_(-a)^(x)f(-t)(-dt)` `=int_(-a)^(x)f(t)dt=int_(-a)^(a)f(t)dt+int_(a)^(x)f(t)dt` `=0+int_(a)^(x)f(t)dt=phi(x)`.
2 Answers 1 views
Let A be an `mxxn` matrix. If there exists a matrix L of type `nxxm` such that `LA=I_(n)`, then L is called left inverse of A. Similarly, if there exi
Correct Answer - C As second row of all the options is same, we have to look at the elements of the first row. Let the left inverse be `[(a,b,c),(d,e,f)]`. Then...
2 Answers 1 views