1.Evaluate the integral. (Use C for the constant of integration.)
integral ln(sqrtx)dx
2. Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the curves about the given axis.
y = 4ex, y = 4e−x, x = 1; about the y-axis
3. Calculate the average value of f(x) = 2x sec2(x) on the interval [0, π/4].
4. Evaluate the integral. (Use C for the constant of integration.)
integral 7 ln(cubertx)dx
integration by parts is just the chain rule in reverse:
d(uv) = u dv + v du
So, you have
#1.
∫ ln √x dx
So, let u = ln √x = 1/2 lnx
dv = dx
Then du = 1/(2x) dx and v = x
∫ ln √x dx = uv - ∫ v du
= x ln √x - ∫ 1/2 dx
= x ln √x - 1/2 x
= 1/2 x (lnx - 1) + C
#2. You have a curved triangular region with vertices at (0,4), (1,4e), (1,4/e)
so, using shells of thickness dx,
v = ∫[0,1] 2πrh dx
where r = x and h = 4e^x - 4e^-x
v = ∫[0,1] 2πx(4e^x - 4e^-x) dx
you can check your answer using discs (washers) of thickness dy, but you have to split the region into two parts because the left boundary changes where the curves intersect at (0,4). You also have to express x as a function of y.
v = ∫[4/e,4] π(R^2-r^2) dy + ∫[4,4e] π(R^2-r^2) dy
where R=1 and r changes from -ln(y/4) to ln(y/4)
v = ∫[4/e,4] π(1-(-ln(y/4))^2) dy + ∫[4,4e] π(1-(ln(y/4))^2) dy
but that's a bit more work ...
#3. the average value of f(x) on [a,b] is (∫[a,b] f(x) dx)/(b-a)That is, the area divided by the width gives the average height. So, in this case, the average is
(∫2x sec^2(x) dx)/(π/4 - 0)
Use integration by parts, with u = 2x and dv = sec^2(x) dx
#4.
∫7 ln∛x dx = 7/3 ∫ lnx dx
See #1.
It just occurred to me that I said chain rule, when it's really the product rule.
Bu I'm sure you caught that.
Looking ahead, differentiation under the integral sign is the chain rule in reverse.
1. To evaluate the integral ∫ ln(√x) dx, we can use the technique of integration by substitution. Let's go through the steps:
Let u = √x, then du/dx = 1/(2√x).
Rearranging, we get dx = 2√x du.
Substituting these values into the integral, we have:
∫ ln(√x) dx = ∫ ln(u) * 2√x du
= 2∫ ln(u) * u du
Now, we can integrate ∫ ln(u) * u du. This is a standard integral that involves integration by parts:
By applying integration by parts, we let dv = ln(u) du and u = u.
Differentiating u, we get du = du.
Integrating dv, we have v = ∫ ln(u) du.
Using the integration formula for ln(u), we get v = u(ln(u) - 1).
Now, applying the integration by parts formula, we have:
∫ ln(u) * u du = u(ln(u) - 1) - ∫(u du)
= u(ln(u) - 1) - ∫ du
= u(ln(u) - 1) - u + C
Finally, substituting back u = √x, we have:
∫ ln(√x) dx = 2(√x)(ln(√x) - 1) - √x + C
Therefore, the evaluated integral is 2√x(ln(√x) - 1) - √x + C.
2. To find the volume V generated by rotating the region bounded by the curves y = 4ex, y = 4e−x, x = 1 about the y-axis, we can use the method of cylindrical shells.
The volume V can be calculated using the following formula:
V = 2π ∫ [x * (f(x) - g(x))] dx,
where f(x) is the top curve, g(x) is the bottom curve, and the integral is taken over the appropriate interval.
In this case, f(x) = 4ex, g(x) = 4e−x, and the interval is from x = 1 to the point of intersection between the two curves.
To find the point of intersection, we set f(x) equal to g(x) and solve for x:
4ex = 4e−x.
Dividing both sides by 4e−x, we have:
ex = e−x.
Taking the natural logarithm of both sides, we get:
x = −x.
Solving for x, we have:
2x = 0.
Hence, x = 0 is the point of intersection.
Substituting this into the volume calculation formula, we have:
V = 2π ∫ [x * (f(x) - g(x))] dx
= 2π ∫ [x * (4ex - 4e−x)] dx
= 8π ∫ [xex - xe−x] dx.
Now, we integrate each term separately.
For ∫ xex dx, we can use integration by parts, with u = x, dv = ex dx:
∫ xex dx = xex - ∫ ex dx
= xex - ex + C1, where C1 is the constant of integration.
For ∫ xe−x dx, we can again use integration by parts, with u = x, dv = e−x dx:
∫ xe−x dx = −xe−x - ∫ −e−x dx
= −xe−x + e−x + C2, where C2 is the constant of integration.
Finally, substituting these results back into the volume calculation formula, we get:
V = 8π [xex - xe−x]
= 8π [(xex - ex) + (ex - xe−x)]
= 8π [(x-1)ex + (1-x)e−x].
Therefore, the volume generated by rotating the region about the y-axis is 8π [(x-1)ex + (1-x)e−x].
3. To calculate the average value of f(x) = 2x sec^2(x) on the interval [0, π/4], we use the following formula:
Average value = (1/(b-a)) * integral from a to b of f(x) dx.
In this case, a = 0 and b = π/4, so the average value can be found using:
Average value = (1/(π/4 - 0)) * integral from 0 to π/4 of 2x sec^2(x) dx.
Simplifying, we have:
Average value = (4/π) * integral from 0 to π/4 of 2x sec^2(x) dx.
To evaluate this integral, we can use integration by parts:
Let u = 2x and dv = sec^2(x) dx.
Then du = 2 dx and v = tan(x).
Applying the integration by parts formula, we have:
Integral of u dv = uv - integral of v du
= 2x tan(x) - ∫ tan(x) * 2 dx
= 2x tan(x) - 2∫ tan(x) dx
Using the integral formula for tan(x), we have:
Integral of tan(x) dx = -ln|cos(x)| + C.
Substituting back into the integration by parts result, we get:
Integral of u dv = 2x tan(x) - 2(-ln|cos(x)|) + C
= 2x tan(x) + 2ln|cos(x)| + C.
Now, substituting this into the average value formula:
Average value = (4/π) * (2x tan(x) + 2ln|cos(x)| + C) evaluated from 0 to π/4
= (4/π) * [2(π/4) tan(π/4) + 2ln|cos(π/4)| - (2(0) tan(0) + 2ln|cos(0)|)]
= (4/π) * [π/2 + 2ln[1/(√2)] - (0 + 2ln(1))]
= (4/π) * [π/2 + 2ln[√2] - 0]
= 2 + 4ln[√2].
Therefore, the average value of f(x) = 2x sec^2(x) on the interval [0, π/4] is 2 + 4ln[√2].
4. To evaluate the integral ∫ 7 ln(∛x) dx, we can follow these steps:
Using the properties of logarithms, we can rewrite the integrand as:
∫ ln(x^(1/3))^7 dx.
By applying the property ln(a^b) = b ln(a), the integral becomes:
∫ 7 ln(x^(1/3)) dx
= ∫ ln(x) dx^(7/3).
Now, let u = x and dv = ln(x) dx^(7/3).
Then du = dx and v = x^(4/3)/(4/3) = (3/4) x^(4/3).
Applying the integration by parts formula, we have:
-7/3 ∫ ln(x) dx^(7/3)
= -7/3 * [x^(4/3) * ln(x) - ∫ (3/4) x^(4/3) (1/x) dx]
= -7/3 * [x^(4/3) * ln(x) - (3/4) ∫ x^(1/3) dx]
= -7/3 * [x^(4/3) * ln(x) - (3/4) * (3/4) * x^(4/3)]
= -7/3 * [x^(4/3) * ln(x) - 9/16 * x^(4/3)] + C,
where C is the constant of integration.
Therefore, the evaluated integral is -7/3 * [x^(4/3) * ln(x) - 9/16 * x^(4/3)] + C.