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Answered on 06 Apr Learn Continuity and Differentiability
Sadika
To verify the Mean Value Theorem (MVT) for the function \( f(x) = x^2 + 2x + 3 \) on the interval \( [4, 6] \), we need to check three conditions:
1. \( f(x) \) is continuous on \( [4, 6] \).
2. \( f(x) \) is differentiable on \( (4, 6) \).
3. There exists a point \( c \) in \( (4, 6) \) such that \( f'(c) = \frac{{f(b) - f(a)}}{{b - a}} \), where \( a = 4 \) and \( b = 6 \).
Let's check these conditions:
1. **Continuity of \( f(x) \) on \( [4, 6] \)**:
The function \( f(x) = x^2 + 2x + 3 \) is a polynomial function and is continuous everywhere. Therefore, it is continuous on the interval \( [4, 6] \).
2. **Differentiability of \( f(x) \) on \( (4, 6) \)**:
The function \( f(x) = x^2 + 2x + 3 \) is a polynomial function and is differentiable everywhere. Therefore, it is differentiable on the interval \( (4, 6) \).
3. **Applying the MVT**:
We need to find \( f'(x) \) and then find a \( c \) such that \( f'(c) = \frac{{f(6) - f(4)}}{{6 - 4}} \).
\[ f'(x) = 2x + 2 \]
Now evaluate \( f'(x) \) at \( x = c \):
\[ f'(c) = 2c + 2 \]
Now evaluate \( f(6) \) and \( f(4) \):
At \( x = 6 \):
\[ f(6) = 6^2 + 2(6) + 3 = 36 + 12 + 3 = 51 \]
At \( x = 4 \):
\[ f(4) = 4^2 + 2(4) + 3 = 16 + 8 + 3 = 27 \]
Now apply MVT condition:
\[ f'(c) = \frac{{f(6) - f(4)}}{{6 - 4}} = \frac{{51 - 27}}{2} = \frac{24}{2} = 12 \]
Now, we need to find \( c \) such that \( 2c + 2 = 12 \):
\[ 2c + 2 = 12 \]
\[ 2c = 12 - 2 \]
\[ 2c = 10 \]
\[ c = 5 \]
So, there exists a point \( c \) in \( (4, 6) \) such that \( f'(c) = \frac{{f(6) - f(4)}}{{6 - 4}} \).
Therefore, the Mean Value Theorem is verified for the function \( f(x) = x^2 + 2x + 3 \) on the interval \( [4, 6] \), and \( c = 5 \) is the point that satisfies the theorem.
Answered on 06 Apr Learn Integrals
Sadika
To evaluate the integral
∫3axb2+c2x2 dx∫b2+c2x23axdx
we can use the substitution method. Let's make the substitution:
u=cxu=cx
Then, du=c dxdu=cdx, or dx=1c dudx=c1du.
Substituting uu and dxdx into the integral:
∫3axb2+c2x2 dx=∫3ab2+u2⋅1c du∫b2+c2x23axdx=∫b2+u23a⋅c1du
Now, we can factor out the constant 3acc3a and rewrite the integral in a more familiar form:
3ac∫1b2+u2 duc3a∫b2+u21du
This is a standard integral, known to be the arctangent function:
3ac⋅1barctan(ub)+Cc3a⋅b1arctan(bu)+C
Now, we need to substitute back u=cxu=cx:
3ac⋅1barctan(cxb)+Cc3a⋅b1arctan(bcx)+C
Thus, the integral evaluates to:
3abarctan(cxb)+Cb3aarctan(bcx)+C
where CC is the constant of integration.
Answered on 06 Apr Learn Integrals
Sadika
To integrate tan(8x)sec4(x)tan(8x)sec4(x) with respect to xx, we can use the substitution method. Let's denote u=tan(x)u=tan(x). Then du=sec2(x) dxdu=sec2(x)dx.
Now, we need to express everything in terms of uu. First, we express tan(8x)tan(8x) in terms of uu:
tan(8x)=sin(8x)cos(8x)=sin(8x)cos4(x)cos(8x)cos4(x)=sin(8x)cos5(x)=2sin(8x)(1+cos(2x))5=2sin(8x)(1+u2)5tan(8x)=cos(8x)sin(8x)=cos4(x)cos(8x)cos4(x)sin(8x)=cos5(x)sin(8x)=(1+cos(2x))52sin(8x)=(1+u2)52sin(8x)
Now, we have u=tan(x)u=tan(x) and du=sec2(x) dxdu=sec2(x)dx. Also, tan(8x)=2sin(8x)(1+u2)5tan(8x)=(1+u2)52sin(8x). So, our integral becomes:
∫2sin(8x)(1+u2)5⋅1sec2(x) du∫(1+u2)52sin(8x)⋅sec2(x)1du
=2∫sin(8x)(1+u2)5 du=2∫(1+u2)5sin(8x)du
Now, we can use a reduction formula to integrate sin(8x)(1+u2)5(1+u2)5sin(8x). Let's denote I(n)I(n) as the integral:
I(n)=∫sin(8x)(1+u2)n duI(n)=∫(1+u2)nsin(8x)du
Then, we have:
I(n)=−1(n−1)(1+u2)n−1+u2(n−1)(n−3)(1+u2)n−3+1(n−1)(n−3)I(n−2)I(n)=−(n−1)(1+u2)n−11+(n−1)(n−3)(1+u2)n−3u2+(n−1)(n−3)1I(n−2)
Now, we apply this reduction formula to our integral:
I(5)=−14(1+u2)4+u24⋅2(1+u2)2+14⋅2I(3)I(5)=−4(1+u2)41+4⋅2(1+u2)2u2+4⋅21I(3)
I(5)=−14(1+u2)4+u28(1+u2)2+18I(3)I(5)=−4(1+u2)41+8(1+u2)2u2+81I(3)
Now, we need to find I(3)I(3):
I(3)=−12(1+u2)2+u22(1+u2)+12I(1)I(3)=−2(1+u2)21+2(1+u2)u2+21I(1)
I(3)=−12(1+u2)2+u22(1+u2)+12I(1)I(3)=−2(1+u2)21+2(1+u2)u2+21I(1)
Now, we need to find I(1)I(1). I(1)I(1) can be directly integrated:
I(1)=−cos(8x)1+u2+arctan(u)+CI(1)=−1+u2cos(8x)+arctan(u)+C
Now, we can substitute I(1)I(1) back into I(3)I(3) and I(3)I(3) back into I(5)I(5). Then, we substitute u=tan(x)u=tan(x) back in terms of xx. We would end up with a long expression involving xx and tan(x)tan(x).
However, please note that the reduction formula involves high-level algebraic manipulations, and it might be very complex to carry out by hand. If you have access to a symbolic computation software like Mathematica or Wolfram Alpha, you can use it to find the antiderivative easily.
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Answered on 06 Apr Learn Differential Equations
Sadika
To determine the order and degree of the differential equation (y′′′)2+(y′′′)3+(y′′)4+y5=0(y′′′)2+(y′′′)3+(y′′)4+y5=0, let's first understand the terminologies:
In the given equation:
Therefore, the differential equation (y′′′)2+(y′′′)3+(y′′)4+y5=0(y′′′)2+(y′′′)3+(y′′)4+y5=0 is a third-order differential equation and has a degree of 5.
Answered on 06 Apr Learn Differential Equations
Sadika
To verify that the function y=acos(x)+bsin(x)y=acos(x)+bsin(x), where a,b∈Ra,b∈R, is a solution of the differential equation d2ydx2+y=0dx2d2y+y=0, we need to substitute this function into the given differential equation and show that it satisfies the equation.
Given function: y=acos(x)+bsin(x)y=acos(x)+bsin(x)
First, let's find the first and second derivatives of yy with respect to xx:
dydx=−asin(x)+bcos(x)dxdy=−asin(x)+bcos(x) d2ydx2=−acos(x)−bsin(x)dx2d2y=−acos(x)−bsin(x)
Now, let's substitute these derivatives into the given differential equation:
d2ydx2+y=(−acos(x)−bsin(x))+(acos(x)+bsin(x))dx2d2y+y=(−acos(x)−bsin(x))+(acos(x)+bsin(x))
=(−acos(x)+acos(x))+(−bsin(x)+bsin(x))=(−acos(x)+acos(x))+(−bsin(x)+bsin(x))
=0=0
Since the expression simplifies to 0, it verifies that the function y=acos(x)+bsin(x)y=acos(x)+bsin(x) is a solution of the differential equation d2ydx2+y=0dx2d2y+y=0.
Answered on 08 Apr Learn Differential Equations
Sadika
To form the differential equation of the family of circles having a center on the y-axis and a radius of 3 units, we first need to express the equation of a circle with its center on the y-axis and radius 3 units.
The general equation of a circle with center (0, c) and radius r is given by:
x2+(y−c)2=r2x2+(y−c)2=r2
In this case, since the center lies on the y-axis, the x-coordinate of the center is 0, and the y-coordinate can be any value (let's denote it as cc).
Substituting cc with yy and rr with 33, we get:
x2+(y−y)2=32x2+(y−y)2=32 x2+y2=9x2+y2=9
Now, to form the differential equation, we'll differentiate both sides with respect to xx:
ddx(x2)+ddx(y2)=ddx(9)dxd(x2)+dxd(y2)=dxd(9) 2x+2ydydx=02x+2ydxdy=0
This can be simplified to:
x+ydydx=0x+ydxdy=0
And finally, solving for dydxdxdy, we get the differential equation:
dydx=−xydxdy=−yx
This is the differential equation of the family of circles with centers on the y-axis and radius 3 units.
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Answered on 08 Apr Learn Differential Equations
Sadika
To find the general solution of the given differential equation:
dydx=1+y21+x2dxdy=1+1+x2y2
Let's rewrite the equation in a more suitable form:
dydx=11+y21+x2dxdy=11+1+x2y2
Now, this looks like a separable differential equation. We can rewrite it as:
dydx=11+y21+x2dxdy=11+1+x2y2 dydx=1+x2+y21+x2dxdy=1+x21+x2+y2
Now, we can separate variables:
1+x21+x2+y2dy=dx1+x2+y21+x2dy=dx
Integrating both sides:
∫1+x21+x2+y2 dy=∫dx∫1+x2+y21+x2dy=∫dx
Let's denote u=1+x2+y2u=1+x2+y2, then du=2y dydu=2ydy:
12∫1u du=x+C21∫u1du=x+C
12ln∣u∣=x+C21ln∣u∣=x+C
Substitute back u=1+x2+y2u=1+x2+y2:
12ln∣1+x2+y2∣=x+C21ln∣1+x2+y2∣=x+C
ln∣1+x2+y2∣=2x+2Cln∣1+x2+y2∣=2x+2C
∣1+x2+y2∣=e2x+2C∣1+x2+y2∣=e2x+2C
1+x2+y2=Ae2x1+x2+y2=Ae2x
Where A=±e2CA=±e2C.
Finally, if we let A=e2CA=e2C, we can express the solution explicitly:
1+x2+y2=e2x1+x2+y2=e2x
y2=e2x−1−x2y2=e2x−1−x2
y=±e2x−1−x2y=±e2x−1−x2
So, the general solution of the given differential equation is:
y=±e2x−1−x2y=±e2x−1−x2
Answered on 08 Apr Learn Differential Equations
Sadika
Answered on 08 Apr Learn Differential Equations
Sadika
Take Class 12 Tuition from the Best Tutors
Answered on 08 Apr Learn Differential Equations
Sadika
To find the integrating factor of the given differential equation (1−x2)dydx−xy=1(1−x2)dxdy−xy=1, we can use the formula for the integrating factor μ(x)μ(x), which is given by:
where P(x)P(x) is the coefficient of yy in the given differential equation.
In this case, P(x)=−xP(x)=−x.
So,
Therefore, the integrating factor μ(x)μ(x) is:
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