Wize University Calculus 1 Textbook > Applications of Integration

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The acceleration due to gravity of an object is given by a(t)=−9.8a\left(t\right)=-9.8a(t)=−9.8. If the initial velocity and displacement are given by v(0)=5v\left(0\right)=5v(0)=5 and d(0)=2d\left(0\right)=2d(0)=2, find the function that represents the displacement at time t.

d(t)=−4.9t2+5t+2d\left(t\right)=-4.9t^2+5t+2d(t)=−4.9t2+5t+2

d(t)=−4.9t2−5t+2d\left(t\right)=-4.9t^2-5t+2d(t)=−4.9t2−5t+2

d(t)=−9.8t+2d\left(t\right)=-9.8t+2d(t)=−9.8t+2

d(t)=−9.8t2+5t+2d\left(t\right)=-9.8t^2+5t+2d(t)=−9.8t2+5t+2

d(t)=−9.8t2−5t+2d\left(t\right)=-9.8t^2-5t+2d(t)=−9.8t2−5t+2

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If the displacement of a particle is given by s(t)=∫0tx2sin⁡xdxs\left(t\right)=\int_0^t\frac{x^2}{\sin x}dxs(t)=∫0t​sinxx2​dx, then the acceleration at at t=π2t=\frac{\pi}{2}t=2π​ is

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Find the average value of the function g(x)=ax+xag\left(x\right)=a^x+x^ag(x)=ax+xa  over the interval [2,3]\left[2,3\right][2,3]  , where a ≠−1a\ \ne-1a =−1  is a constant.

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Find the average value of y=x2y=x^2y=x2 over [1,3][1, 3][1,3] .

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Find the average value of f(x)=sin⁡xf(x)=\sin xf(x)=sinx on [0,π/2][0,\pi/2][0,π/2].

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Find the area of the region bounded between x2−x+1x^2-x+1x2−x+1 and the line y=1y=1y=1.

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Find integral that represents the area enclosed by the curves y=x3y=x^3y=x3 and y=xy=xy=x.

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 Find the area bounded by the curves  y=−x  and  y= x2 −2.y=-x\ \text{ and }\ y=\ x^2\ -2.y=−x  and  y= x2 −2.

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Find the area bounded by the curves y=cos⁡xy=\cos xy=cosx and y=sin⁡xy=\sin xy=sinx from 0 to π/2\pi/2π/2. Sketch the area.

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Find the volume of the solid that is produced by revolving the region bounded between x=1+y2x=1+y^2x=1+y2, x=0x=0x=0, y=1y=1y=1, and y=2y=2y=2 about the xxx-axis.

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Find the volume of the solid that is produced by revolving y=xsin⁡xcos⁡xy=\sqrt{x\sin x\cos x}y=xsinxcosx​ about the xxx-axis, between 0 and π/2\pi/2π/2.

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Find the volume of the solid that is produced by revolving the region bounded between y=x2/4y=x^2/4y=x2/4, y=0y=0y=0, and x=2x=2x=2 about the yyy-axis. Use the method of cylindrical shells.

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Write the integrals representing the volume of the solid that is produced by revolving the region bounded between y=tan⁡3xy=\tan^3xy=tan3x, y=1y=1y=1, and x=0x=0x=0 about the line y=2y=2y=2 using both the washer method and the method of cylindrical shells. Do not compute the integrals.

Washer: V=π∫0π/4(2−tan⁡3x)2−(2−1)2dxV = \pi\displaystyle\int^{\pi/4}_0(2 − \tan^3x)^2 − (2 − 1)^2 dxV=π∫0π/4​(2−tan3x)2−(2−1)2dx

Shells: V=2π∫01(2−y)arctan⁡(y3)dyV = 2\pi\displaystyle\int^1_0(2 − y)\arctan(\sqrt[3]{y}) dyV=2π∫01​(2−y)arctan(3y​)dy

Washer: V=π∫0π/4(2+tan⁡3x)2−(2−1)2dxV=\pi\displaystyle\int_0^{\pi/4}(2+\tan^3x)^2−(2−1)^2dxV=π∫0π/4​(2+tan3x)2−(2−1)2dx

Shells: V=2π∫01(2−y)arctan⁡(y3)dyV=2\pi\displaystyle\int_0^1(2−y)\arctan(\sqrt[3]{y})dyV=2π∫01​(2−y)arctan(3y​)dy

Washer: V=π∫0π/4(2+tan⁡3x)2−(2−1)2dxV=\pi\displaystyle\int_0^{\pi/4}(2+\tan^3x)^2−(2−1)^2dxV=π∫0π/4​(2+tan3x)2−(2−1)2dx

Shells: V=2π∫01(2+y)arctan⁡(y3)dyV=2\pi\displaystyle\int_0^1(2+y)\arctan(\sqrt[3]{y})dyV=2π∫01​(2+y)arctan(3y​)dy

Washer: V=π∫0π/4(2−tan⁡3x)2−(2−1)2dxV=\pi\displaystyle\int_0^{\pi/4}(2−\tan^3x)^2−(2−1)^2dxV=π∫0π/4​(2−tan3x)2−(2−1)2dx

Shells: V=2π∫01(2+y)arctan⁡(y3)dyV=2\pi\displaystyle\int_0^1(2+y)\arctan(\sqrt[3]{y})dyV=2π∫01​(2+y)arctan(3y​)dy

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Find the arc length function for the curve y=2x3/2y=2x^{3/2}y=2x3/2 with starting point P0=(0,0)P_0=\left(0,0\right)P0​=(0,0).

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Practice Question
Find the arc length function for the curve y=2x32y=2x^{\frac{3}{2}}y=2x23​ with starting point P0(0,3)P_0\left(0,3\right)P0​(0,3).

Arc length

s(x)=

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Find the arc length of y=exy=e^xy=ex on [0,1][0,1][0,1].

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Q:\textbf{Q:}Q: Consider the curve  y=f(x)=x33+14xy=f(x)=\dfrac{x^3}{3}+\dfrac{1}{4x}y=f(x)=3x3​+4x1​. Find the arc length of this curve over the interval [1,3][1, 3][1,3]. Find a function giving the arc length of the curve over the interval [1,x].[1, x].[1,x].

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Set up the appropriate integral for determining the surface area of revolution for the following curve: y=x2\bcb{y = x^2}y=x2, (0≤x≤2\bcb{0 \leq x \leq 2}0≤x≤2), about the y\bcb{y}y-axis.

2π∫02x1+4x2  ⁣dx.\displaystyle { 2\pi\int_0^{2} x\sqrt{1+4x^2}  \de{x}}.2π∫02​x1+4x2​ dx.

2π∫02x1+2x2  ⁣dx.\displaystyle { 2\pi\int_0^{2} x\sqrt{1+2x^2}  \de{x}}.2π∫02​x1+2x2​ dx.

2π∫02x1+2x  ⁣dx.\displaystyle { 2\pi\int_0^{2} x\sqrt{1+2x}  \de{x}}.2π∫02​x1+2x​ dx.

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Find the area of the surface obtained by revolving the curve 
y=9−x2y = \sqrt{9 -x^2}y=9−x2​
from x=−3x = -3x=−3 to x=3x = 3x=3 about the xxx - axis.

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