- What is,
lim n--> ∞ nΣ(k=1) (2+k*(5/n))^3 * 5/n
as a definite integral.
- Solve d/dt x^4∫2 tan (x^2)dx
- What is the area bounded by the curves x = y^2 - 4y and x = 2y - y^2?
for the limit, I get
∫[2,7] x^3 dx
For the derivative of the integral, I get
tan((x^4)^2)*(4x^3)
For the area, the two curves intersect at (0,0) and (-3,3). So the area is
∫[0,3] (2y-y^2)-(y^2-4y) dy = 9
To find the limit in the first question, we can rewrite the expression using the definition of a definite integral. Recall that ∫[a, b] f(x) dx represents the area under the curve of f(x) from x = a to x = b.
So, let's rewrite the given expression as a definite integral:
lim n--> ∞ nΣ(k=1) (2+k*(5/n))^3 * 5/n
= lim n--> ∞ [ Σ(k=1) (2+k*(5/n))^3 * 5/n ] * n
= lim n--> ∞ [ Σ(k=1) (2+k*(5/n))^3 * (5/n) ] * (5/n) * n
= lim n--> ∞ [ Σ(k=1) (2+k*(5/n))^3 * (5/n) ] * ∆x
where ∆x = 5/n and represents the width of each rectangle.
Now, we can rewrite the expression as a definite integral:
lim n--> ∞ [ ∑(k=1) (2+k*(5/n))^3 * (5/n) ] * ∆x
= lim n--> ∞ [ ∑(k=1) (2+k*(∆x))^3 * ∆x ] * ∆x
= lim n--> ∞ [ ∑(k=1) (2+k*(∆x))^3 * ∆x ] * dx
where dx represents an infinitesimally small change in x.
Finally, as n approaches infinity, ∆x approaches 0 and the sum becomes an integral:
lim n--> ∞ [ ∑(k=1) (2+k*(∆x))^3 * ∆x ] * dx
= ∫[1, ∞] (2+k*dx)^3 * dx
Therefore, the expression lim n--> ∞ nΣ(k=1) (2+k*(5/n))^3 * 5/n can be represented as the definite integral ∫[1, ∞] (2+k*dx)^3 * dx.
Moving on to the second question:
To solve d/dt x^4∫2 tan (x^2)dx, we need to use the Leibniz's Rule for Differentiation under the Integral Sign.
Let's simplify the expression first:
d/dt x^4∫2 tan (x^2)dx
= ∫2 d/dt (x^4 tan (x^2)) dx
= ∫2 (4x^3 tan (x^2) + x^4 d(tan(x^2))/dt) dx
Next, we need to find d(tan(x^2))/dt. Since tan(x^2) is a function of x, we will consider x^2 as a variable u and rewrite the expression as tan(u) instead:
d(tan(u))/dt
= d(tan(u))/du * du/dt
= sec^2(u) * 0 (since du/dt = 0 for constant u)
= 0
Hence, d(tan(x^2))/dt = 0, and it does not affect the differentiation.
So, the original expression becomes:
∫2 (4x^3 tan (x^2)) dx
Now, we can integrate term by term:
= 4∫2 (x^3 tan (x^2)) dx
To evaluate this integral, we might need to use integration by substitution or other techniques. However, since the integral is not the main focus of the question, we won't go through the specific steps for integration.
Finally, in the third question:
To find the area bounded by the curves x = y^2 - 4y and x = 2y - y^2, we need to determine the points of intersection between these two curves.
Setting the two equations equal to each other:
y^2 - 4y = 2y - y^2
2y^2 - 6y = 0
Factorizing, we get:
2y(y - 3) = 0
So, the solutions are y = 0 and y = 3.
Now, we integrate the difference of the two curves with respect to y, from y = 0 to y = 3:
Area = ∫[0, 3] (2y - y^2) - (y^2 - 4y) dy
= ∫[0, 3] 2y - y^2 - y^2 + 4y dy
= ∫[0, 3] 6y - 2y^2 dy
To find the definite integral, we integrate term by term:
Area = [3y^2/2 - (2/3)y^3] evaluated from 0 to 3
Substituting the limits:
Area = [ (3(3)^2/2 - (2/3)(3)^3 ) ] - [ (3(0)^2/2 - (2/3)(0)^3) ]
= [ (3(9)/2 - (2/3)(27) ) ] - [ (0/2 - 0) ]
= [ (27/2 - 18) ] - [ (0 - 0) ]
= [ (27/2 - 36/2) ] - [ (0) ]
= -9/2
Therefore, the area bounded by the curves x = y^2 - 4y and x = 2y - y^2 is -9/2.