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  • Math
    To draw an angle of 105˚ with a compass. First draw a horizontal ray, from 1 endpoint going to the east. Then using the center of the protractor, place the center on the endpoint. The compass will...

    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

    2 educator answers.

  • 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.

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

    Asked by enotes on via web

    1 educator answer.

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