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Question 4 Find the value of the summation. ∑ k=1 3 ​ 2 k = ∫ −4 3 ​ (6x 5 +2x 4 )dx Question 9 Evaluate the definite integral. ∫ 0 4 π ​ ​ sin(4t)dt ∫(5x 4 +2x 2 )dx
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Question 4 Find the value of the summation. ∑

k=1
3
​
2
k
= ∫
−4
3
​
(6x
5
+2x
4
)dx Question 9 Evaluate the definite integral. ∫
0
4
π
​

​
sin(4t)dt ∫(5x
4
+2x
2
)dx

User Alemjerus
by
7.8k points

1 Answer

5 votes

Answer:

sum_(k=1)^3 2^k

Explanation:

Its easy. ;)

User Linusg
by
8.4k points
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