
MathSturges' rule is a way of calculating the number of bins (e.g. categories or classes) of a set of data. It is assumed that the data come from a normally distributed population. Sturges' rule states...

MathWe are given that the population mean mu=12.074, with a population standard deviation sigma=.046. We are asked to find the percentage of the population with the following properties: (a) Find...

MathA pareto chart consists of a bar chart, with bars in descending order of length, and a line graph representing the cumulative total. Please see the attached graph:

MathWhat is the probability of 45 children in a class having their birthday on the same day of the week?Probability is the measure of the likelihood that an event occurs. The problem requires you to determine what the probability is of 45 children being born on the same day of the week. There are 7...

Math1. Construction of angle 105 degree using compass:(Refer attached image) Steps: Draw a line AB and mark point O on it where angle is to be drawn. With O as center draw an arc (semicircle) which...

MathYou need to obtain the reciprocal of the number `(5x),` hence, you need to flip the number over, or to divide 1 by the number `(5x), ` such that: `1/(5x)` Using the properties of multiplication...

MathHello! If we could split boys and girls into n teams, and b boys and g girls will be in each team, then obviously `nb=48` and `ng=60.` Therefore `n` divides 48 and `n` divides 60, or the same may...

MathYou need to use the substitution `2y = u` , such that: `2y= u => 2dy = du => dy= (du)/(2)` Replacing the variable, yields: `int y*e^(2y) dy = (1/4)int u*e^u du` You need to use the...

Math`int_1^sqrt(3)arctan(1/x)dx` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)dx=f(x)g(x)intf'(x)g(x)dx` If we write f(x)=u and g'(x)=v, then `intuvdx=uintvdxint(u'intvdx)dx`...

Math`int_0^(1/2)cos^(1)xdx` Let's first evaluate the indefinite integral by using the method of integration by parts, `intcos^(1)xdx=cos^(1)x*int1dxint(d/dx(cos^(1)x)int1dx)dx`...

Math`int_1^2(ln(x))^2/x^3dx` If f(x) and g(x) are differentiable function, then `intf(x)g'(x)=f(x)g(x)intf'(x)g(x)dx` If we rewrite f(x)=u and g'(x)=v, them `intuvdx=uintvdxint(u'intvdx)dx` Using the...

MathYou need to use the substitution `sin x = t,` such that: `sin x = t => cos x dx = dt` Replacing the variable, yields: `int cos x*ln(sin x) dx = int ln t dt` You need to use the integration by...

Math`int_0^1r^3/sqrt(4+r^2)dr` Let's first evaluate the indefinite integral using the method of substitution, Substitute `x=4+r^2, =>r^2=x4` `=>dx=2rdr`...

Math`int_1^2x^4(ln(x))^2dx` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)dx=f(x)g(x)intf'(x)g(x)dx` If we write f(x)=u and ` ` g'(x)=v, then `intuvdx=uintvdxint(u'intvdx)dx` Using...

MathYou need to use integration by parts, such that: `int udv = uv  int vdu` `u = e^s => du = e^s ds` `dv = sin(ts) => v = (cos(ts))/(1)` `int e^s sin (ts) ds = e^s*cos(ts)  int...

MathYou need to use the following substitution, such that: `sqrt x = t => (dx)/(2sqrt x) = dt => dx = 2tdt` Replacing the variable yields: `int cos sqrt x dx = int (cos t)*(2tdt)` You need to...

Math`intt^3e^(t^2)dt` Let `x=t^2` `dx=2tdt` `intt^3e^(t^2)dt=intxe^(x)dx/2` `=1/2intxe^(x)dx` Now apply integration by parts, If f(x) and g(x) are differentiable functions then,...

MathYou need to use the substitution `theta^2= t` , such that: `theta^2 = t => 2theta d theta= dt => theta d theta= (dt)/2` Replacing the variable, yields: `int_(sqrt(pi/2))^(sqrt pi)...

MathYou need to use the substitution `cos t = u` , such that: `cos t = u => sin t dt = du => sin t dt = du` Replacing the variable, yields: `int_0^pi e^(cos t)*sin (2t)dt = 2int_0^pi e^(cos...

MathYou need to use the substitution `1+x = t` , such that: `1+x = t => dx = dt` Changing the variable yields: `int x*ln(1+x) dx = int (t1)*ln t dt = int t*ln t dt  int ln t dt` You need to...

MathYou need to use integration by parts, such that: `int udv = uv  int v du` `u = sin(ln x) => du = cos(ln x)(1/x) dx` `dv = 1 => v = x` `int sin(ln x) dx = x*sin(ln x)  int (cos(ln x))/x...

MathYou need to use the substitution `beta*t = u` , such that: `beta*t = u => beta dt = du ` Replacing the variable, yields: `int t^2*sin(beta*t) dt = 1/(beta^3) int u^2*sin u du` You need to use...

Math`int ln (root(3)(x)) dx` To evaluate, apply integration by parts `int udv = uv  int vdu` . So let `u = ln (root(3)(x))=ln (x^(1/3))=1/3ln(x)` and `dv = dx` Then, differentiate u and integrate...

Math`intsin^1xdx` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)dx=f(x)g(x)intf'(x)g(x)dx` If we write f(x)=u and g'(x)=v, then `intuvdx=uintvdxint(u'intvdx)dx` Using the above...

Math`intarctan(4t)dt` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)dx=f(x)g(x)intf'(x)g(x)dx` If we write f(x)=u and g'(x)=v, then `intuvdx=uintvdxint(u'intvdx)dx` Using the above...

Math`int p^5 ln(p) dp` To evaluate, apply integration by parts intu dv = uv int vdu. So let `u= ln (p)` and `dv = p^5 dp` Then, differentiate u and integrate dv. `du=1/p dp` and `v = int p^5dp =...

Math`inttsec^2(2t)dt` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)dx=f(x)g(x)intf'(x)g(x)dx` If we write f(x)=u and g'(x)=v, then `intuvdx=uintvdxint(u'intvdx)dx` Now using the...

MathYou need to use the integration by parts such that: `int fdg = fg  int gdf` `f = s => df = ds` `dg = 2^s=> g = (2^s)/(ln 2)` `int s*2^s ds = s* (2^s)/(ln 2)  int (2^s)/(ln 2) ds` `int s*2^s...

Math`int (ln x)^2dx` To evaluate, apply integration by parts `int udv= uv  vdu` . So let `u = (lnx)^2` and `dv = dx` Then, differentiate u and integrate dv. `du = 2lnx * 1/x dx = (2lnx)/x dx` and `v =...

MathTo help you solve this, we consider the the integration by parts: `int u * dv = uv  int v* du` Let `u = t` and `dv = sinh(mt) dt.` based from `int t*sinh(mt) dt` for` int u*dv` In this integral,...

Math`inte^2thetasin(3theta)d theta` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)=f(x)g(x)intf'(x)g(x)dx` If we rewrite f(x)=u and g'(x)=v, then `intuvdx=uintvdxint(u'intvdx)dx`...

MathYou need to use integration by parts, such that: `int udv = uv  int vdu` `u = e^(theta) => du = e^(theta)d theta` `dv = cos (2theta) => v = (sin 2 theta)/2` `int e^(theta)cos (2theta) d...

Math`intz^3e^zdz` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)=f(x)g(x)intf'(x)g(x)dx` If we write f(x)=u and g'(x)=v, then `intuvdx=uintvdxint(u'intvdx)dx` Using the above...

Math`intxtan^2xdx` Rewrite the integrand using the identity `tan^2x=sec^2x1` `intxtan^2xdx=intx(sec^2x1)dx` `=intxsec^2xdxintxdx` Now let's evaluate `intxsec^2xdx` using integration by parts,...

Math`int(xe^(2x))/(1+2x)^2dx` If f(x) and g(x) are differentiable functions, then `intf(x)g'(x)=f(x)g(x)intf'(x)g(x)dx` If we rewrite f(x)=u and g'(x)=v, then `intuvdx=uintvdxint(u'intvdx)dx` Using...

MathYou need to use integration by parts, such that: `int udv = uv  int v du` `u = (arcsin^2(x))=> du = (2arcsin x)/(sqrt(1x^2))dx` `dv = 1 => v = x` `int (arcsin^2(x)) dx = x*(arcsin^2(x)) ...

MathYou need to solve the integral `int_0^(1/2) (x) cos (pi*x) dx` , hence, you need to use substitution `pi*x = t => pi*dx = dt => dx = (dt)/(pi)` `int x*cos (pi*x) dx = 1/(pi^2) int t*cos t`...

Math`int_0^1(x^2+1)e^xdx` Let's first evaluate the indefinite integral,using the method of integration by parts `int(x^2+1)e^xdx=(x^2+1)inte^xdxint(d/dx(x^2+1)inte^xdx)dx`...

MathTake the indefinite integral by parts. It is known (and easy to prove) that `cosh(t)=d/dt sinh(t)` and `sinh(t)= d/dt cosh(t).` Denote `u=t,` `v=sinh(t),` then `du=dt` and `dv=cosh(t)dt.` `int t...

Math`int_4^9 lny/sqrty dy` To evaluate, apply integration by parts `intudv=uvintvdu` . So let: `u = ln y ` and `dv=int1/sqrty dy` Then, differentiate u and integrate dv. `du=1/y dy` and `v=int...

Math`int_1^3 r^3 ln(r) dr` To evaluate, apply integration by parts `int udv = uv  vdu` . So let `u = ln r` and `dv = r^3 dr` Then, differentiate u and integrate dv. `u=1/r dr` and `v= int r^3...

MathYou need to use the integration by parts for `int_0^(2pi) t^2*sin(2t)dt` such that: `int udv = uv  int vdu` `u = t^2 => du = 2tdt` `dv = sin 2t=> v =(cos 2t)/2` `int_0^(2pi) t^2*sin(2t)dt...

Math`intxcos(5x)dx` To evaluate, apply integration by parts `int udv = uv intv du` . So let: `u=x ` and `dv =intcos(5x)dx` Differentiating u and integrating dv yield: `du=dx ` and...

MathYou need to use the substitution `0.2y = u` , such that: `0.2y= u => 0.2dy = du => dy= (du)/(0.2)` Replacing the variable, yields: `int y*e^(0.2y) dy = (10/2)int u/(0.2)*e^u du` You need to...

MathYou need to use the substitution `3t = u` , such that: `3t= u => 3dt = du => dt = (du)/3` Replacing the variable, yields: `int t*e^(3t) dt = (1/9)int u*e^u du` You need to use the...

MathYou need to solve the integral `int (x1) sin (pi*x) dx = int x*sin (pi*x) dx  int sin (pi*x)dx` You need to use substitution `pi*x = t => pi*dx = dt => dx = (dt)/(pi)` `int x*sin (pi*x) dx...

Math`int (x^2+2x)cosx dx` To evaluate, apply integration by parts `int udv = uv  int vdu` . So let `u = x^2+2x` and `dv = cosx dx` Then, differentiate u and integrate dv. `du = (2x + 2)dx` and `v =...

MathThe average value is the integral of a function over an interval divided by the length of an interval. The length here is 2, let's found the integral: `int_(1)^1 1/(32u) du = 1/2...

MathThe average value is the integral of a function over an interval divided by the length of an interval. The length here is 3, found the integral: `int_2^5 (x3)^2 dx = 1/3 (x3)^3_(x=2)^5 =...

MathThe average value on an interval is the integral on this interval divided by the length of the interval. The length is 2. The integral is `ln(x)_(x=1)^3=ln3ln1=ln3.` So the average value is...
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