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Post a LessonAnswered on 06 Apr Learn Unit III: Calculus/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 Unit III: Calculus/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 Unit III: Calculus/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.
Answered on 06 Apr Learn Unit III: Calculus/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 Unit III: Calculus/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.
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