• WILD SWANS: THREE DAUGHTERS OF CHINA

Who are the Three Daughters of China? Jung Chang, Er-Hong, and who?

Jung Chang’s grandmother, Yu-fang, was the first in the generation of three daughters. This woman experienced severe hardship and turmoil at an early age. When she was two, her feet were bound to…

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• HISTORY

Which metal replaced bronze?

Historians and archaeologists generally agree that iron replaced bronze as the metal of choice for crafting tools. Bronze is composed of copper and tin, both of which are quite soft metals. A tool…

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• THE MERCHANT OF VENICE

In William Shakespeare’s The Merchant of Venice, why do Bassanio and Antonio go only to Shyolck…

If, as has been suggested, William Shakespeare adapted his play The Merchant of Venice from that of his prematurely-deceased contemporary Christopher Marlowe’s The Jew of Malta (also known more…

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• THE KITE RUNNER

The strong underlying force of this novel is the relationship between Amir and Hassan. Discuss…

While Amir and Hassan grow up as close as brothers, differences in ethnic group, religious belief, and economic status cause rifts between them. Their relationship is preceded by that of their…

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• HISTORY

Why did dynasties form in Sumer?

Sumer’s location as the southernmost region of Mesopotamia, along with its access to the Persian Gulf, made it a prime location for a blossoming of culture and economy. Many believe that the…

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• GEORGE ORWELL

What are some instances in which George Orwell demonstrated leadership?

Two of George Orwell’s most famous novels, Animal Farm and 1984, demonstrate a healthy dose of leadership. However, in standard Orwellian fashion (his own name being from which the term derives),…

• ALL MY SONS

What is Miller’s criticism on the American dream in All My Sons?

Arthur Miller is making a statement about achieving the American dream through Joe Keller, who will do anything to obtain it. Joe, a manufacturer of airplane cylinder heads, knowingly ships out…

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• THE SCARLET LETTER

What reasons are given for Hester staying in Boston in The Scarlett Letter?

The narrator tells us in the final chapter that “Here had been her sin; here, her sorrow; and here was yet to be her penitence.” In other words, Hester had chosen to return to Boston because this…

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• SCIENCE

What is an explanation of Planck’s Constant in layman’s terms?

Max Planck was a physicist who lived during the early 20th century. He concluded that the transfer of energy must work differently at the subatomic level. At the subatomic level, things are VERY…

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• THE SCARLET LETTER

In The Scarlet Letter, what reasons are given for Hester staying in Boston?

In The Scarlet Letter, Hester Prynne chooses to stay in Boston out of attachment to the life-changing events that had happened there and to the person involved in them–the man who desecrated her…

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• THE ODYSSEY

What laws of behavior and attitude does Polyphemus violate in his treatment of the Greeks in The…

The most obvious law of behavior that Polyphemus violates is hospitality. In Ancient Greece, and much of the rest of the ancient world, hospitality was much more important. Journeys took much…

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• TO KILL A MOCKINGBIRD

In To Kill a Mockingbird, was Boo Radley mad when his brother put cement into the tree? And why…

First, I want to be clear that it was Nathan Radley, Boo’s father, who put cement in the knot of the tree outside of the Radley home. Although he claimed to Jem and Scout that this was because the…

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• MATH

∫sin(ln(x))dx∫sin(ln(x))dx First make a substitution and then use integration by parts to evaluate…

We need to make a substitution then use integration by parts. Let us make the substitution: ln(x)=t,ln(x)=t, so: x=etx=et therefore dx=etdtdx=etdt so our equation can be changed.

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• MATH

∫xln(1+x)dx∫xln(1+x)dx First make a substitution and then use integration by parts to evaluate…

You need to use the substitution 1+x=t1+x=t , such that: 1+x=t⇒dx=dt1+x=t⇒dx=dt Changing the variable yields: ∫x⋅ln(1+x)dx=∫(t−1)⋅lntdt=∫t⋅lntdt−∫lntdt∫x⋅ln(1+x)dx=∫(t-1)⋅lntdt=∫t⋅lntdt-∫lntdtYou need to…

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• MATH

∫π0ecos(t)sin(2t)dt∫0πecos(t)sin(2t)dt First make a substitution and then use integration by parts…

You need to use the substitution cost=ucost=u , such that: cost=u⇒−sintdt=du⇒sintdt=−ducost=u⇒-sintdt=du⇒sintdt=-du Replacing the variable, yields:

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• MATH

∫π√π2√θ3cos(θ2)dθ∫π2πθ3cos(θ2)dθ First make a substitution and then…

You need to use the substitution θ2=tθ2=t , such that: θ2=t⇒2θdθ=dt⇒θdθ=dt2θ2=t⇒2θdθ=dt⇒θdθ=dt2 Replacing the variable, yields:

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• MATH

∫t3e−t2dt∫t3e-t2dt First make a substitution and then use integration by parts to…

∫t3e−t2dt∫t3e-t2dt Let x=t2x=t2 dx=2tdtdx=2tdt ∫t3e−t2dt=∫xe−xdx2∫t3e-t2dt=∫xe-xdx2 =12∫xe−xdx=12∫xe-xdx Now apply integration by parts, If f(x) and g(x) are differentiable functions then,…

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• MATH

∫cos(x−−√)dx∫cos(x)dx First make a substitution and then use integration by parts to…

You need to use the following substitution, such that: x−−√=t⇒dx2x−−√=dt⇒dx=2tdtx=t⇒dx2x=dt⇒dx=2tdt Replacing the variable yields: ∫cosx−−√dx=∫(cost)⋅(2tdt)∫cosxdx=∫(cost)⋅(2tdt) You need to…

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• MATH

∫t0essin(t−s)ds∫0tessin(t-s)ds Evaluate the integral

You need to use integration by parts, such that: ∫udv=uv−∫vdu∫udv=uv-∫vduu=es⇒du=esdsu=es⇒du=esds dv=sin(t−s)⇒v=−cos(t−s)−1dv=sin(t-s)⇒v=-cos(t-s)-1

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• MATH

∫21×4(ln(x))2dx∫12×4(ln(x))2dx Evaluate the integral

∫21×4(ln(x))2dx∫12×4(ln(x))2dx If f(x) and g(x) are differentiable functions, then ∫f(x)g'(x)dx=f(x)g(x)−∫f'(x)g(x)dx∫f(x)g′(x)dx=f(x)g(x)-∫f′(x)g(x)dx If we write f(x)=u and  g'(x)=v, then ∫uvdx=u∫vdx−∫(u’∫vdx)dx∫uvdx=u∫vdx-∫(u′∫vdx)dx Using…

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• MATH

∫10r34+r2−−−−−√dr∫01r34+r2dr Evaluate the integral

∫10r34+r2−−−−−√dr∫01r34+r2dr Let’s first evaluate the indefinite integral using the method of substitution, Substitute x=4+r2,⇒r2=x−4x=4+r2,⇒r2=x-4 ⇒dx=2rdr⇒dx=2rdr…

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• MATH

∫cos(x)ln(sin(x))dx∫cos(x)ln(sin(x))dx Evaluate the integral

You need to use the substitution sinx=t,sinx=t, such that: sinx=t⇒cosxdx=dtsinx=t⇒cosxdx=dt Replacing the variable, yields: ∫cosx⋅ln(sinx)dx=∫lntdt∫cosx⋅ln(sinx)dx=∫lntdt You need to use the integration by…

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• MATH

∫21(ln(x))2x3dx∫12(ln(x))2x3dx Evaluate the integral

∫21(ln(x))2x3dx∫12(ln(x))2x3dx If f(x) and g(x) are differentiable function, then ∫f(x)g'(x)=f(x)g(x)−∫f'(x)g(x)dx∫f(x)g′(x)=f(x)g(x)-∫f′(x)g(x)dx If we rewrite f(x)=u and g'(x)=v, them ∫uvdx=u∫vdx−∫(u’∫vdx)dx∫uvdx=u∫vdx-∫(u′∫vdx)dx Using the…

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• MATH

∫120cos−1xdx∫012cos-1xdx Evaluate the integral

∫120cos−1xdx∫012cos-1xdx Let’s first evaluate the indefinite integral by using the method of integration by parts, ∫cos−1xdx=cos−1x⋅∫1dx−∫(ddx(cos−1x)∫1dx)dx∫cos-1xdx=cos-1x⋅∫1dx-∫(ddx(cos-1x)∫1dx)dx…

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• MATH

∫3√1arctan(1x)dx∫13arctan(1x)dx Evaluate the integral

∫3√1arctan(1x)dx∫13arctan(1x)dx If f(x) and g(x) are differentiable functions, then ∫f(x)g'(x)dx=f(x)g(x)−∫f'(x)g(x)dx∫f(x)g′(x)dx=f(x)g(x)-∫f′(x)g(x)dx If we write f(x)=u and g'(x)=v, then ∫uvdx=u∫vdx−∫(u’∫vdx)dx∫uvdx=u∫vdx-∫(u′∫vdx)dx…

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• MATH

∫ye2ydy∫ye2ydy Evaluate the integral

You need to use the substitution −2y=u-2y=u , such that: −2y=u⇒−2dy=du⇒dy=−du2-2y=u⇒-2dy=du⇒dy=-du2 Replacing the variable, yields: ∫y⋅e−2ydy=(14)∫u⋅eudu∫y⋅e-2ydy=(14)∫u⋅eudu You need to use the…

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• MATH

∫2π0t2sin(2t)dt∫02πt2sin(2t)dt Evaluate the integral

You need to use the integration by parts for ∫2π0t2⋅sin(2t)dt∫02πt2⋅sin(2t)dt such that: ∫udv=uv−∫vdu∫udv=uv-∫vdu u=t2⇒du=2tdtu=t2⇒du=2tdt dv=sin2t⇒v=−cos2t2dv=sin2t⇒v=-cos2t2

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• MATH

∫31r3ln(r)dr∫13r3ln(r)dr Evaluate the integral

∫31r3ln(r)dr∫13r3ln(r)dr To evaluate, apply integration by parts ∫udv=uv−vdu∫udv=uv-vdu . So let u=lnru=lnr and dv=r3drdv=r3dr Then, differentiate u and integrate dv. u=1rdru=1rdr and

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• MATH

∫94ln(y)y√dy∫49ln(y)ydy Evaluate the integral

∫94lnyy√dy∫49lnyydy To evaluate, apply integration by parts ∫udv=uv−∫vdu∫udv=uv-∫vdu . So let: u=lnyu=lny and dv=∫1y√dydv=∫1ydy Then, differentiate u and integrate dv. du=1ydydu=1ydy and

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• MATH

∫10tcosh(t)dt∫01tcosh(t)dt Evaluate the integral

Take the indefinite integral by parts. It is known (and easy to prove) that cosh(t)=ddtsinh(t)cosh(t)=ddtsinh(t) and sinh(t)=ddtcosh(t).sinh(t)=ddtcosh(t). Denote u=t,u=t, v=sinh(t),v=sinh(t), then du=dtdu=dt and dv=cosh(t)dt.dv=cosh(t)dt.

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• MATH

∫10(x2+1)e−xdx∫01(x2+1)e-xdx Evaluate the integral

∫10(x2+1)e−xdx∫01(x2+1)e-xdx Let’s first evaluate the indefinite integral,using the method of integration by parts ∫(x2+1)e−xdx=(x2+1)∫e−xdx−∫(ddx(x2+1)∫e−xdx)dx∫(x2+1)e-xdx=(x2+1)∫e-xdx-∫(ddx(x2+1)∫e-xdx)dx…

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• MATH

∫120xcos(πx)dx∫012xcos(πx)dx Evaluate the integral

You need to solve the integral ∫120(x)cos(π⋅x)dx∫012(x)cos(π⋅x)dx , hence, you need to use substitution π⋅x=t⇒π⋅dx=dt⇒dx=dtππ⋅x=t⇒π⋅dx=dt⇒dx=dtπ ∫x⋅cos(π⋅x)dx=1π2∫t⋅cost∫x⋅cos(π⋅x)dx=1π2∫t⋅cost…

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• MATH

∫(arcsin(x))2dx∫(arcsin(x))2dx Evaluate the integral

You need to use integration by parts, such that: ∫udv=uv−∫vdu∫udv=uv-∫vduu=(arcsin2(x))⇒du=2arcsinx1−x2−−−−−√dxu=(arcsin2(x))⇒du=2arcsinx1-x2dx dv=1⇒v=xdv=1⇒v=x

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• MATH

∫xe2x(1+2x)2dx∫xe2x(1+2x)2dx Evaluate the integral

∫xe2x(1+2x)2dx∫xe2x(1+2x)2dx If f(x) and g(x) are differentiable functions, then ∫f(x)g'(x)=f(x)g(x)−∫f'(x)g(x)dx∫f(x)g′(x)=f(x)g(x)-∫f′(x)g(x)dx If we rewrite f(x)=u and g'(x)=v, then ∫uvdx=u∫vdx−∫(u’∫vdx)dx∫uvdx=u∫vdx-∫(u′∫vdx)dx Using…

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• MATH

∫xtan2xdx∫xtan2xdx Evaluate the integral

∫xtan2xdx∫xtan2xdx Rewrite the integrand using the identity tan2x=sec2x−1tan2x=sec2x-1 ∫xtan2xdx=∫x(sec2x−1)dx∫xtan2xdx=∫x(sec2x-1)dx =∫xsec2xdx−∫xdx=∫xsec2xdx-∫xdx Now let’s evaluate ∫xsec2xdx∫xsec2xdx using integration by parts,…

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• MATH

∫z3ezdz∫z3ezdz Evaluate the integral

∫z3ezdz∫z3ezdz If f(x) and g(x) are differentiable functions, then ∫f(x)g'(x)=f(x)g(x)−∫f'(x)g(x)dx∫f(x)g′(x)=f(x)g(x)-∫f′(x)g(x)dx If we write f(x)=u and g'(x)=v, then ∫uvdx=u∫vdx−∫(u’∫vdx)dx∫uvdx=u∫vdx-∫(u′∫vdx)dx Using the above…

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• MATH

∫e−θcos(2θ)dθ∫e-θcos(2θ)dθ Evaluate the integral

We have to find the integral ∫e−θcos(2θ)dθ∫e-θcos(2θ)dθ We can do this by integration by parts i.e.

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• MATH

∫e2θsin(3θ)dθ∫e2θsin(3θ)dθ Evaluate the integral

∫e2θsin(3θ)dθ∫e2θsin(3θ)dθ If f(x) and g(x) are differentiable functions, then ∫f(x)g'(x)=f(x)g(x)−∫f'(x)g(x)dx∫f(x)g′(x)=f(x)g(x)-∫f′(x)g(x)dx If we rewrite f(x)=u and g'(x)=v, then ∫uvdx=u∫vdx−∫(u’∫vdx)dx∫uvdx=u∫vdx-∫(u′∫vdx)dx…

1 educator answer

• MATH

∫tsinh(mt)dt∫tsinh(mt)dt Evaluate the integral

To help you solve this, we consider the the integration by parts: ∫u⋅dv=uv−∫v⋅du∫u⋅dv=uv-∫v⋅du Let u=tu=t and dv=sinh(mt)dt.dv=sinh(mt)dt. based from ∫t⋅sinh(mt)dt∫t⋅sinh(mt)dt for∫u⋅dv∫u⋅dv In this integral,…

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• MATH

∫(lnx)2dx∫(lnx)2dx Evaluate the integral

∫(lnx)2dx∫(lnx)2dx To evaluate, apply integration by parts ∫udv=uv−vdu∫udv=uv-vdu . So let u=(lnx)2u=(lnx)2 and dv=dxdv=dx Then, differentiate u and integrate dv. du=2lnx⋅1xdx=2lnxxdxdu=2lnx⋅1xdx=2lnxxdx and

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• MATH

∫s2sds∫s2sds Evaluate the integral

You need to use the integration by parts such that: ∫fdg=fg−∫gdf∫fdg=fg-∫gdf f=s⇒df=dsf=s⇒df=ds dg=2s⇒g=2sln2dg=2s⇒g=2sln2 ∫s⋅2sds=s⋅2sln2−∫2sln2ds∫s⋅2sds=s⋅2sln2-∫2sln2ds

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• MATH

∫tsec22tdt∫tsec22tdt Evaluate the integral

∫tsec2(2t)dt∫tsec2(2t)dt If f(x) and g(x) are differentiable functions, then ∫f(x)g'(x)dx=f(x)g(x)−∫f'(x)g(x)dx∫f(x)g′(x)dx=f(x)g(x)-∫f′(x)g(x)dx If we write f(x)=u and g'(x)=v, then ∫uvdx=u∫vdx−∫(u’∫vdx)dx∫uvdx=u∫vdx-∫(u′∫vdx)dx Now using the…

1 educator answer

• MATH

∫p5lnpdp∫p5lnpdp Evaluate the integral

∫p5ln(p)dp∫p5ln(p)dp To evaluate, apply integration by parts intu dv = uv -int vdu. So let u=ln(p)u=ln(p) and dv=p5dpdv=p5dp Then, differentiate u and integrate dv. du=1pdpdu=1pdp and

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• MATH

∫arctan(4t)dt∫arctan(4t)dt Evaluate the integral

∫arctan(4t)dt∫arctan(4t)dt If f(x) and g(x) are differentiable functions, then ∫f(x)g'(x)dx=f(x)g(x)−∫f'(x)g(x)dx∫f(x)g′(x)dx=f(x)g(x)-∫f′(x)g(x)dx If we write f(x)=u and g'(x)=v, then ∫uvdx=u∫vdx−∫(u’∫vdx)dx∫uvdx=u∫vdx-∫(u′∫vdx)dx Using the above…

1 educator answer

• MATH

∫sin−1xdx∫sin-1xdx Evaluate the integral

∫sin−1xdx∫sin-1xdx If f(x) and g(x) are differentiable functions, then ∫f(x)g'(x)dx=f(x)g(x)−∫f'(x)g(x)dx∫f(x)g′(x)dx=f(x)g(x)-∫f′(x)g(x)dx If we write f(x)=u and g'(x)=v, then ∫uvdx=u∫vdx−∫(u’∫vdx)dx∫uvdx=u∫vdx-∫(u′∫vdx)dx Using the above…

1 educator answer

• MATH

∫ln(x−−√3)dx∫ln(x3)dx Evaluate the integral

∫ln(x−−√3)dx∫ln(x3)dx To evaluate, apply integration by parts ∫udv=uv−∫vdu∫udv=uv-∫vdu . So let u=ln(x−−√3)=ln(x13)=13ln(x)u=ln(x3)=ln(x13)=13ln(x)and dv=dxdv=dx Then, differentiate u and integrate…

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• MATH

∫t2sin(βt)dt∫t2sin(βt)dt Evaluate the integral

You need to use the substitution β⋅t=uβ⋅t=u , such that: β⋅t=u⇒βdt=duβ⋅t=u⇒βdt=du Replacing the variable, yields: ∫t2⋅sin(β⋅t)dt=1β3∫u2⋅sinudu∫t2⋅sin(β⋅t)dt=1β3∫u2⋅sinudu You need to use…

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• MATH

∫(x2+2x)cos(x)dx∫(x2+2x)cos(x)dx Evaluate the integral

∫(x2+2x)cosxdx∫(x2+2x)cosxdx To evaluate, apply integration by parts ∫udv=uv−∫vdu∫udv=uv-∫vdu . So let u=x2+2xu=x2+2x and dv=cosxdxdv=cosxdxThen, differentiate u and integrate dv. du=(2x+2)dxdu=(2x+2)dx and

1 educator answer

• MATH

∫(x−1)sin(πx)dx∫(x-1)sin(πx)dx Evaluate the integral

You need to solve the integral ∫(x−1)sin(π⋅x)dx=∫x⋅sin(π⋅x)dx−∫sin(π⋅x)dx∫(x-1)sin(π⋅x)dx=∫x⋅sin(π⋅x)dx-∫sin(π⋅x)dxYou need to use substitution π⋅x=t⇒π⋅dx=dt⇒dx=dtππ⋅x=t⇒π⋅dx=dt⇒dx=dtπ

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• MATH

∫te−3tdt∫te-3tdt Evaluate the integral

You need to use the substitution -3t=u3t=u , such that: −3t=u⇒−3dt=du⇒dt=−du3-3t=u⇒-3dt=du⇒dt=-du3 Replacing the variable, yields: ∫t⋅e−3tdt=(19)∫u⋅eudu∫t⋅e-3tdt=(19)∫u⋅eudu You need to use the…

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