∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
----
∫(t = 2 to 3) t √(t - 2) dt
= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2
= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du
= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}
= 26/15.
----
For the k-entry, use integration by parts with
u = t, dv = sin(πt) dt
du = 1 dt, v = (-1/π) cos(πt).
So, ∫(t = 2 to 3) t sin(πt) dt
= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt
= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]
= 5/π + 0
= 5/π.
Therefore,
∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.
Answer:
28 degrees
Step-by-step explanation:
If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.The corresponding angles are the ones at the same location at each intersection.
In the figure ABD and EDF are corresponding angles. So they are equal
So equation the angle ABD and EDF , we get
(3x+4) = (7x-20)
Group the like terms,
3x-7x = -20-4
-4x = -24

x = 6
Thus BCD will be,
(6x - 8)
=>(6(6)-8)
=>(36-8)
=> 28 degrees
Answer:
200
Step-by-step explanation:
Substitute 15 for y




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