Hi, I have 3 problems here need to be solved, help me please
1.Evaluate ∫ sin2x sec^5(2x)dx
2. If f(x)=√x and P is a partition of [1,16] determined by 1,3,5,7,9,16,find a Riemann sum Rp of f by choosing the numbers 1,4,5,9 and 9(yes 9 again) in the subintervals of P
3. Given G(x)=∫(limit 1/x to x) √t^4+t^2+4 dt, find G'(x)
A square root is equal to 1/2...
=t^4(1/2)+t^2(1/2)+4^(1/2)......the square root is removed
=t^2+t+2
g'(t)=2t+1
sin u sec^5 u
= tan u sec^4 u
= sec u * tan u * sec^3 u du
if v = sec u, then that is v^3 dv
So, you have
∫ sin2x sec^5(2x)dx
1/2 ∫ sin2x sec^5(2x) 2 dx
= 1/2 ∫ sec^3(2x) d(sec(2x))
= 1/8 sec^4(2x) + C
There are lots of good Riemann sum calculators online.
Using the 2nd Fundamental Theorem of Calculus, G'(x) =
[√(x^4+x^2+4) * (1)] - [√(1/x^4 + 1/x^2 + 4) * (-1/x^2)]
= √(x^4+x^2+4) + √(4x^4+x^2+1)/x^4
Of course! I'd be happy to help you solve these problems.
1. To evaluate the integral ∫ sin^2(x) sec^5(2x) dx, you can use integration by substitution. Let's set u = 2x. Then, the integral becomes ∫ sin^2(x) sec^5(u) (1/2) du.
Next, we can use the trigonometric identity sin^2(x) = 1 - cos^2(x) to rewrite the integral as ∫ (1 - cos^2(x)) sec^5(u) (1/2) du.
Now, we can split this integral into two parts. The first part is ∫ sec^5(u) (1/2) du, which can be integrated using integration by parts. By letting v = sec^4(u) and du = sec(u) tan(u) du, we can find the integral as (1/2) [ sec^4(u) tan(u) - ∫ 4 sec^3(u) tan^2(u) du].
The second part is ∫ cos^2(x) sec^5(u) (1/2) du, which simplifies to ∫ cos^2(x) sec^5(u) (1/2) du.
2. To find the Riemann sum Rp of f(x) = √x on the partition P = {1, 3, 5, 7, 9, 16} using the chosen subintervals with boundaries {1, 4, 5, 9, 9, 16}, we can evaluate f(x) at each of the chosen subinterval endpoints and multiply them by the corresponding subinterval length.
The Riemann sum Rp is given by Rp = ∑[f(c_i)(x_i - x_i-1)], where i ranges from 1 to the number of subintervals (in this case, 5), c_i is any point within the ith subinterval, and x_i and x_i-1 are the upper and lower boundaries of the ith subinterval.
Using the given numbers, we have:
R1 = f(c1)(x1 - x0) = f(1)(4 - 1),
R2 = f(c2)(x2 - x1) = f(4)(5 - 4),
R3 = f(c3)(x3 - x2) = f(5)(9 - 5),
R4 = f(c4)(x4 - x3) = f(9)(9 - 9),
R5 = f(c5)(x5 - x4) = f(9)(16 - 9),
where f(x) = √x.
3. To find G'(x) given G(x) = ∫(1/x to x) √(t^4 + t^2 + 4) dt, we can use the Fundamental Theorem of Calculus. According to this theorem, if F(x) is an antiderivative of a function f(x) over an interval [a, b], then ∫(a to b) f(x) dx = F(b) - F(a).
In this case, G(x) represents the integral of the function √(t^4 + t^2 + 4) from the lower limit of 1/x to the upper limit of x. The expression √(t^4 + t^2 + 4) is the integrand.
To find G'(x), we need to differentiate G(x) with respect to x. By applying the Fundamental Theorem of Calculus, we have G'(x) = (√(x^4 + x^2 + 4)) - (√(1/x^4 + 1/x^2 + 4/x^2)).
To simplify further, you can write G'(x) as (√(x^4 + x^2 + 4)) - (√((1 + x^2 + 4x^2)/x^2)).
I hope this helps you solve the problems! If you have any further questions, please let me know.