Learn Unit III: Calculus with Free Lessons & Tips

To find the antiderivative of the function 3x2+4x33x2+4x3, we'll integrate each term separately using the power rule of integration.

1. For 3x23x2: ∫3x2 dx=3⋅x33+C=x3+C∫3x2dx=3⋅3x3+C=x3+C

2. For 4x34x3: ∫4x3 dx=4⋅x44+C=x4+C∫4x3dx=4⋅4x4+C=x4+C

Putting them together, the antiderivative of 3x2+4x33x2+4x3 is: ∫(3x2+4x3) dx=x3+x4+C∫(3x2+4x3)dx=x3+x4+C

where CC is the constant of integration.

To find the antiderivative F(x)F(x) of the function f(x)=4x3−6f(x)=4x3−6, we integrate each term separately using the power rule of integration.

For 4x34x3, we integrate term-by-term:

∫4x3 dx=4⋅x44+C1=x4+C1∫4x3dx=4⋅4x4+C1=x4+C1

For −6−6, we integrate it as a constant:

∫(−6) dx=−6x+C2∫(−6)dx=−6x+C2

Now, combining the results, we get:

F(x)=x4−6x+CF(x)=x4−6x+C

where C=C1+C2C=C1+C2 is the constant of integration.

Given that F(0)=3F(0)=3, we can find the value of the constant CC:

F(0)=04−6(0)+C=3F(0)=04−6(0)+C=3 C=3C=3

Therefore, the antiderivative F(x)F(x) is:

F(x)=x4−6x+3F(x)=x4−6x+3

To integrate the function 2xsin⁡(x2+1)2xsin(x2+1) with respect to xx using integration by substitution, let's make the substitution:

u=x2+1u=x2+1

Then, we find dudu:

dudx=2xdxdu=2x

Now, we can express dxdx in terms of dudu:

dx=du2xdx=2xdu

Substituting uu and dxdx into the integral, we have:

∫2xsin⁡(x2+1) dx=∫sin⁡(u)⋅du∫2xsin(x2+1)dx=∫sin(u)⋅du

Now, we integrate sin⁡(u)sin(u) with respect to uu:

∫sin⁡(u) du=−cos⁡(u)+C∫sin(u)du=−cos(u)+C

Now, we need to substitute back u=x2+1u=x2+1:

−cos⁡(x2+1)+C−cos(x2+1)+C

So, the antiderivative of 2xsin⁡(x2+1)2xsin(x2+1) with respect to xx is:

−cos⁡(x2+1)+C−cos(x2+1)+C

where CC is the constant of integration.

To form the differential equation representing the family of curves y=asin⁡(x+b)y=asin(x+b), where aa and bb are arbitrary constants, we need to find the first and second derivatives of yy with respect to xx.

Given function: y=asin⁡(x+b)y=asin(x+b)

First, let's find the first derivative of yy with respect to xx:

dydx=acos⁡(x+b)dxdy=acos(x+b)

Now, let's find the second derivative of yy with respect to xx:

d2ydx2=−asin⁡(x+b)dx2d2y=−asin(x+b)

Now, we need to express yy, dydxdxdy, and d2ydx2dx2d2y in terms of yy and xx to form the differential equation.

We have:

y=asin⁡(x+b)y=asin(x+b)

dydx=acos⁡(x+b)dxdy=acos(x+b)

d2ydx2=−asin⁡(x+b)dx2d2y=−asin(x+b)

Substituting these into the differential equation d2ydx2+y=0dx2d2y+y=0, we get:

−asin⁡(x+b)+asin⁡(x+b)=0−asin(x+b)+asin(x+b)=0

0=00=0

This equation is trivially satisfied.

However, let's represent the differential equation without cancelling out terms:

d2ydx2+y=0dx2d2y+y=0

−asin⁡(x+b)+asin⁡(x+b)+asin⁡(x+b)=0−asin(x+b)+asin(x+b)+asin(x+b)=0

asin⁡(x+b)−asin⁡(x+b)=0asin(x+b)−asin(x+b)=0

0=00=0

This differential equation represents the family of curves y=asin⁡(x+b)y=asin(x+b), where aa and bb are arbitrary constants.

The family of lines passing through the origin has the general form y=mxy=mx, where mm is the slope of the line. To find the differential equation representing this family of lines, we need to express the derivative of yy with respect to xx in terms of yy and xx, which will give us the differential equation.

Given the family of lines y=mxy=mx, where mm is the slope, we differentiate yy with respect to xx:

dydx=mdxdy=m

This derivative represents the slope of the line at any point (x,y)(x,y). Now, since y=mxy=mx, we have m=yxm=xy.

Substituting this into dydx=mdxdy=m, we get:

dydx=yxdxdy=xy

This is the differential equation representing the family of lines passing through the origin.