Step-by-step explanation:
A left Riemann sum approximates a definite integral as:
Given ∫₂⁸ cos(x²) dx:
a = 2, b = 8, and f(x) = cos(x²)
Therefore, Δx = 6/n and x = 2 + (6/n) (k − 1).
Plugging into the sum:
∑₁ⁿ cos((2 + (6/n) (k − 1))²) (6/n)
Therefore, the answer is C. Notice that answer D would be a right Riemann sum rather than a left (uses k instead of k−1).
The properties of a vector space are satisfied Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes aren't legitimate are if
Property three does now no longer follow: Suppose that Property three is legitimate, shall we namethe neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently
= 0 If O is the neuter, then it ought to restore x², but This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant
have additive inverse
Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.
Property 5(b) isn't valid: we are able to introduce
a counter example. we could use z = 1 then
v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3
Since 3x ^ 2 +1 ne x^ 2 +3. then the associativity rule doesnt hold.
Property f doesnt observe because of the switch of variables. for instance, if v = x ^ 2 1 × v=1^ × x ^ 2 = 1 #V
Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes arent legitimate.
Step-with the aid of using-step explanation:
Note that each sum and scalar multiplication entails in replacing the order from that most important coefficient with the impartial time period earlier than doing the same old sum/scalar multiplication.
Property three does now no longer follow: Suppose that Property three is legitimate, shall we name v = a × x ^ 2 +bx +c the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently
= zero If O is the neuter, then it ought to restore x², but zero This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant have additive inverse
Let
Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.
Property 5(b) isn't valid: we are able to introduce
a counter example. we could usethen the associativity rule doesnt hold.
Note that each expressions are same because of the distributive rule of actual numbers. Also, you could be aware that his assets holds due to the fact in each instances we 'switch variables twice.
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