# Math Homework Help

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• Math
You need to use the mean value thorem to verify the given inequality, such that: int_a^b f(x)dx = (b-a)*f(c), c in (a,b) Replacing cos x for f(x) and pi/6 for a, pi/4 for b, yields:...

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• Math
`int_-3^0(1+sqrt(9-x^2))dx` Consider the graph of y=f(x)=`1+sqrt(9-x^2)` `y=1+sqrt(9-x^2)` `y-1=sqrt(9-x^2)` `(y-1)^2=9-x^2` `x^2+(y-1)^2=3^2` This is the equation of circle of radius 3 centred at...

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• Math
`int_-5^5(x-sqrt(25-x^2))dx` `=int_-5^5xdx-int_-5^5sqrt(25-x^2)dx` `=I_1-I_2` I_1 can be be interpreted as area of two triangles;one above the x-axis and the other below axis.Since they are on the...

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• Math
`int_(-1)^2 |x| dx` To interpret this in terms of area, graph the integrand. The integrand is the function f(x) =|x|. Then, shade the region bounded by the graph of f(x)=|x| and the x-axis in the...

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• Math
`int_0^10 |x-5|dx` To interpret this in terms of area, graph the integrand. The integrand is the function f(x) = |x - 5|. Then, shade the region bounded by f(x) = |x-5| and the x-axis in the...

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• Math
You need to use the mean value thorem to verify the given inequality, such that: `int_a^b f(x)dx = (b-a)*f(c), c in (a,b)` Replacing `x^2 - 4x + 4` for `f(x)` and 0 for a, 4 for b, yields: `int_0^4...

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• Math
You need to check if` int_0^1 sqrt(1+x^2)dx <= int_0^1sqrt(1+x)dx` , using mean value theorem, such that: `int_a^b f(x)dx = (b-a)f(c), ` where `c in (a,b)` `int_0^1 sqrt(1+x^2)dx <=...

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• Math
You need to use the midpoint rule to approximate the interval. First, you need to find `Delta x` , such that: `Delta x = (b-a)/n` The problem provides b=8, a=0 and n = 4, such that: `Delta x =...

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• Math
You need to use the midpoint rule to approximate the interval. First, you need to find `Delta x,` such that: `Delta x = (b-a)/n` The problem provides `b=pi/2` , a=0 and n = 4, such that: `Delta x =...

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• Math
You need to use the midpoint rule to approximate the interval. First, you need to find `Delta x` , such that: `Delta x = (b-a)/n` The problem provides b=2, a=0 and n = 5, such that: `Delta x =...

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• Math
You need to evaluate the definite integral using the mid point rule, hence, first you need to evaluate `Delta x:` `Delta x = (b-a)/n => Delta x = (5-1)/4 = 1` You need to denote each of the 4...

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• Math
You need to use the fundamental theorem of calculus, to prove the equality, such that: `int_a^b f(x)dx = F(b) - F(a)` You need to replace x for f(x), such that: `int_a^b xdx = x^2/2|_a^b` `int_a^b...

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• Math
You need to evaluate the definite integral, such that: `int_a^b f(x) dx = F(b) - F(a)` `int_a^b x^2 dx = (x^3)/3|_a^b` `int_a^b x^2 dx = (b^3)/3 - (a^3)/3` `int_a^b x^2 dx = (b^3 - a^3)/3` Hence,...

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• Math
You have to recall the definition of the Reiman Integral `int_a^bf(x)dx=lim_(n->oo)sum_(i=1)^nf(x(i))Deltax` `where Deltax =(b-a)/n and x(i)= a +iDeltax` `x ` `a=2 and b = 6 ` `Deltax = (6-2)/n=...

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• Math
`int_1^10 (x-4ln(x))dx` To express this definite integral as limit of Riemann's Sum, apply the formula: `int_a^b f(x) dx = lim_(n-> oo)sum_(i=1)^oo f(x_i)Delta x` where `Delta x = (b-a)/n` `x_i...

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• Math
`int_-1^2(1-x)dx` To interpret the integral in terms of area , graph the integrand. The integrand is the function `f(x)=1-x` Graph the function in the interval (-1,2). Refer the attached graph. The...

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• Math
`int_0^9 (1/3x-2)dx` To interpret this integral in terms of area, graph the integrand. The integrand is the function `f(x)=1/3x-2` . Then, shade the region bounded by `f(x)=1/3x-2` and the...

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• Math
`v(t)=2t-1/(1+t^2)` position of the particle s(t) is given by, `s(t)=intv(t)dt` `s(t)=int(2t-1/(1+t^2))dt` `s(t)=2(t^2/2)-arctan(t)+C` , C is constant `s(t)=t^2-arctan(t)+C` Now let's find C ,...

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• Math
`a(t) = sin(t) + 3cos(t)` `v(t) = -cos(t) + 3sin(t) + a` `Now, v(0) = 2` `Thus, 2 = -cos(0)+3sin(0)+a` `or, 2 = -1 + 0 + a` `or., a = 3` `Now, v(t) = -cos(t) + 3sin(t) + 3` `Thus, s(t) = -sin(t) -...

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• Math
You need to evaluate the asimptotes of the function, such that: `lim_(x->-pi/2,x>-pi/2)(4x - tan x) = -2pi - tan(-pi/2) = 2pi + tan (pi/2) = oo` `lim_(x->pi/2,x<pi/2)(4x - tan x) = 2pi...

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• Math
`y=(x-2)e^(-x)` (I) Asymptotes To determine its horizontal asymptotes, take the limit of this function as x approaches positive and negative infinity. `lim_(x->-oo) (x-2)e^(-x) =-oo`...

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• Math
`y=f(x)=x+ln(x^2+1)` a) Asymptotes The function has no undefined points , so it has no vertical asymptotes. For Horozontal asymptotes , check if at `x->+-oo` the function behaves as a line...

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• Math
`f'(x) = cos(x) - (1-x^2)^(-1/2)` `f(x) = sin(x) - sin^-1(x) + c` ``

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• Math
`f'(x) = 2e^x + sec(x)tan(x)` `f(x) = 2e^x + sec(x)` ``

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• Math
`f'(x)=sqrt(x^3)+root(3)(x^2)` `f(x)=intf'(x)dx` `f(x)=int(sqrt(x^3)+root(3)(x^2))dx` apply the sum rule, `f(x)=intsqrt(x^3)dx+introot(3)(x^2)dx` `f(x)=x^(3/2+1)/(3/2+1)+x^(2/3+1)/(2/3+1)+C` , C is...

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• Math
You need to evaluate the function f using the provided information, hence, you need to apply the antiderivative, such that: `int (f'(x) sinhx + 2cosh x)dx = int (f'(x)(e^x - e^(-x))/2) + (e^x +...

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• Math
`f'(t)=2t-3sin(t)` `f(t)=int(2t-3sin(t))dt` `f(t)=2(t^2/2)-3(-cos(t))+C` ,C is constant `f(t)=t^2+3cos(t)+C` Now , evaluate C , given f(0)=5 `f(0)=5=0^2+3cos(0)+C` `5=3+C` `C=2`...

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• Math
You need to evaluate f(u) using the antiderivative of the function f'(u), such that: `int f'(u) du = f(u) + c` `int (u^2 + sqrt u)/u du = int (u^2)/u du + int (sqrt u)/u du` `int (u^2 + sqrt u)/u...

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• Math
`f''(x) = 1 - 6x + 48x^2` `f'(x) = x - 3x^2 + 16x^3 + a` `Now, f'(0) = 2` `i.e. 2 = a` `Thus, f'(x) = x - 3x^2 + 16x^3 + 2` `f(x) = (x^2/2) - x^3 + 4x^4 + 2x + b` `Now, f(0) = 1` `Thus, b = 1`...

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• Math
`f''(x) = 2x^3 + 3x^2 - 4x + 5` `or, f'(x) = {(1/2)x^4} + x^3 - 2x^2 + 5x + a` `or, f(x) = {(1/10)*x^5} + (x^4/4) - {(2/3)x^3} + {(5/2)x^2} + ax + b` `Now, f(0) = 2` `i.e. 2 = b` `Also, f(1) = 0`...

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• Math
`lim_(x->oo)(x^2-x^3)e^(2x)` `=lim_(x->oo)x^3(1/x-1)e^(2x)` Apply the Algebraic limit theorem, `=lim_(x->oo)x^3(lim_(x->oo)1/x-1)lim_(x->oo)e^(2x)` plug in the value of x...

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• Math
You need to evaluate the limit, hence, you need to replace` pi` for x in expression under limit: `lim_(x->pi)(x-pi)csc x = (pi-pi)/(sin pi) = 0/0` Since the limit is indeterminate `0/0` , you...

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• Math
`lim_(x->1^+)(x/(x-1)-1/ln(x))` `=lim_(x->1^+)(xln(x)-1(x-1))/((x-1)ln(x))` `=lim_(x->1^+)(xln(x)-x+1)/((x-1)ln(x))` Apply L'Hospital's rule, Test L'Hospital condition :0/0...

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• Math
`lim_(x->(pi/2)^-)(tan(x))^cos(x)` `=lim_(x->(pi/2)^-)e^(cos(x)lntan(x))` applying the limit chain rule, `lim_(x->(pi/2)^-)cos(x)lntan(x)` `=lim_(x->(pi/2)^-)(lntan(x)/(1/cos(x)))`...

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• Math
You need to evaluate the horizontal asymptotes to the graph of function, such that: `lim_(x->+oo) 2-2x-x^3 = oo` `lim_(x->-oo) 2-2x-x^3 = oo` Notice that the graph has no vertical or...

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• Math
`y=x^3-6x^2-15x+4` a) Asymptotes Polynomial function of degree 1 or higher can't have any asymptotes. b) Maxima/Minima `y'=3x^2-12x-15` `y'=3(x^2-4x-5)` `y'=3(x-5)(x+1)` Now we can find critical...

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• Math
The function has no vertical, horizontal or slant asymptotes. You need to determine the extrema of the function, hence, you need to determine the zeroes of the first derivative: `f'(x) =...

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• Math
`y=x/(1-x^2)` a) Asymptotes Vertical asymptotes are the zeros of the denominator of the function. `1-x^2=0rArr(1+x)(1-x)=0` `x=-1 , x=1` Vertical asymptotes are x=1 and x=-1 Degree of numerator=1...

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• Math
`y=1/(x(x-3)^2)` a) Asymptotes Vertical asymptotes are the zeros of the denominator. `x(x-3)^2=0rArr x=0,x=3` So, x=0 and x=3 are the vertical asymptotes. Degree of numerator=0 Degree of...

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• Math
`y=1/x^2-1/(x-2)^2` To determine the asymptotes, express the function as a single fraction. Since the LCD of the two fractions is x^2(x-2)^2, then, the function becomes: `y = 1/x^2 *...

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• Math
`y=x^2/(x+8)` a) Asymptotes Vertical asymptotes are the zeros of the denominator `x+8=0rArrx=-8` Vertical asymptote is x=-8 Degree of numerator=2 Degree of denominator=1 Degree of...

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• Math
`y=sqrt(1-x)+sqrt(1+x)` Domain of function: `-1<=x<=1` a) Asymptotes a) Asymptotes The function has no undefined points, so there are no vertical asymptotes. For horizontal asymptotes , check...

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• Math
`y=xsqrt(2+x)` a) Asymptotes Domain of function is x `>=` -2 Since the function has no undefined points , so it has no vertical asymptotes. Horizontal Asymptotes: Let's find the limits of the...

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• Math
`y=root(3)(x^2+1)` a) Asymptotes Since the function has no undefined point, so it has no vertical asymptote. For horizontal asymptotes check if at x`->+-oo` , the function behaves as a line...

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• Math
You need to evaluate the local absolute extrema of the function, hence, you need to find the zeroes of the equation f'(x) = 0. You need to evaluate the derivative such that: `f'(x) = 3x^2 - 12x +...

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• Math
You need to evaluate the local absolute extrema of the function, hence, you need to find the zeroes of the equation f'(x) = 0. You need to evaluate the derivative using product rule, such that:...

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• Math
You need to evaluate the local absolute extrema of the function, hence, you need to find the zeroes of the equation f'(x) = 0. You need to evaluate the derivative using the quotient rule: `f'(x) =...

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• Math
`f(x)=sqrt(x^2+x+1)` Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the...

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• Math
You need to evaluate the extreme values of the function on the interval [-pi,pi], hence, you need to evaluate the zeroes of the function f'(x). You need to find the derivative of the function:...

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