∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
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∫(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.
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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>.
so,
it equals 12.49 or 12.5 depending on rounding
I hope this helps!
The interest is simple
A=5,200×(1+0.04×5)
A=6,240
180-104=76
76x2=152
Answer is 152
Answer:
c
Step-by-step explanation:
i took a test on it
