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  • Math
    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...

    Asked by tonys538 on via web

    1 educator answer.

  • Math
    1. 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...

    Asked by saiabh360 on via web

    1 educator answer.

  • Math
    You 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...

    Asked by fluberdoodlez on via web

    1 educator answer.

  • Math
    Hello! 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...

    Asked by gramma635 on via web

    1 educator answer.

  • Math
    You 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...

    Asked by enotes on via web

    1 educator answer.

  • 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=uintvdx-int(u'intvdx)dx`...

    Asked by enotes on via web

    1 educator answer.

  • 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*int1dx-int(d/dx(cos^(-1)x)int1dx)dx`...

    Asked by enotes on via web

    1 educator answer.

  • 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=uintvdx-int(u'intvdx)dx` Using the...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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...

    Asked by enotes on via web

    1 educator answer.

  • 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=x-4` `=>dx=2rdr`...

    Asked by enotes on via web

    1 educator answer.

  • 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=uintvdx-int(u'intvdx)dx` Using...

    Asked by enotes on via web

    1 educator answer.

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

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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...

    Asked by enotes on via web

    1 educator answer.

  • 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,...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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)...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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 (t-1)*ln t dt = int t*ln t dt - int ln t dt` You need to...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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...

    Asked by enotes on via web

    1 educator answer.

  • 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...

    Asked by enotes on via web

    1 educator answer.

  • 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=uintvdx-int(u'intvdx)dx` Using the above...

    Asked by enotes on via web

    1 educator answer.

  • 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=uintvdx-int(u'intvdx)dx` Using the above...

    Asked by enotes on via web

    1 educator answer.

  • 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 =...

    Asked by enotes on via web

    1 educator answer.

  • 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=uintvdx-int(u'intvdx)dx` Now using the...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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...

    Asked by enotes on via web

    1 educator answer.

  • 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 =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    To 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,...

    Asked by enotes on via web

    1 educator answer.

  • 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=uintvdx-int(u'intvdx)dx`...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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...

    Asked by enotes on via web

    1 educator answer.

  • 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=uintvdx-int(u'intvdx)dx` Using the above...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `intxtan^2xdx` Rewrite the integrand using the identity `tan^2x=sec^2x-1` `intxtan^2xdx=intx(sec^2x-1)dx` `=intxsec^2xdx-intxdx` Now let's evaluate `intxsec^2xdx` using integration by parts,...

    Asked by enotes on via web

    1 educator answer.

  • 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=uintvdx-int(u'intvdx)dx` Using...

    Asked by enotes on via web

    1 educator answer.

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

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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`...

    Asked by enotes on via web

    1 educator answer.

  • 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^-xdx-int(d/dx(x^2+1)inte^-xdx)dx`...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Take 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...

    Asked by enotes on via web

    1 educator answer.

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

    Asked by enotes on via web

    1 educator answer.

  • 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...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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...

    Asked by enotes on via web

    1 educator answer.

  • 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...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You 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...

    Asked by enotes on via web

    2 educator answers.

  • Math
    You 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...

    Asked by enotes on via web

    1 educator answer.

  • Math
    You need to solve the integral `int (x-1) 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...

    Asked by enotes on via web

    1 educator answer.

  • 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 =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    The 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/(3-2u) du = -1/2...

    Asked by enotes on via web

    1 educator answer.

  • Math
    The 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 (x-3)^2 dx = 1/3 (x-3)^3|_(x=2)^5 =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    The 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=ln3-ln1=ln3.` So the average value is...

    Asked by enotes on via web

    1 educator answer.

  • Math
    The average value on an interval is the integral on this interval divided by the length of the interval. The length is `pi` . The integral is `(-2cos(x)+1/2 cos(2x))|_0^pi=2+1/2+2-1/2=4.` So the...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Given `f(x)=(2x)/(1+x^2)^2, [0, 2]` Average Value Formula=`1/(b-a)int_a^bf(x)dx` `f_(ave)=1/(2-0)int_0^2(2x)/(1+x^2)^2dx` `=(1/2)*2int_0^2x/(1+x^2)^2dx` `=int_0^2x/(1+x^2)^2dx` Integrate using...

    Asked by enotes on via web

    1 educator answer.

  • Math
    The average value on an interval is the integral on this interval divided by the length of the interval. The length is 4. The integral is `(2x^2- 1/3 x^3)|_(x=0)^(4)=32-4*16/3.` So the average...

    Asked by enotes on via web

    1 educator answer.

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