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Sergeu [11.5K]
2 years ago
8

Find the most general antiderivative of the function. (check your answer by differentiation. use c for the constant of the antid

erivative. ) g(t) = 8 t t2 t
Mathematics
1 answer:
katovenus [111]2 years ago
8 0

The most general antiderivative of the given function g(t) is (8t + t³/3 + t²/2 + c).

The antiderivative of a function is the inverse function of a derivative.

This inverse function of the derivative is called integration.

Here the given function is: g(t) = 8 + t² + t

Therefore, the antiderivative of the given function is

∫g(t) dt

= ∫(8 + t² + t) dt

= ∫8 dt + ∫t² dt + ∫t dt

= [8t⁽⁰⁺¹⁾/(0+1) + t⁽²⁺¹⁾/(2+1) + t⁽¹⁺¹⁾/(1+1) + c]

= (8t + t³/3 + t²/2 + c)

Here 'c' is the constant.

Again, differentiating the result, we get:

d/dt(8t + t³/3 + t²/2 + c)

= [8 ˣ 1 ˣ t⁽¹⁻¹⁾ + 3 ˣ t⁽³⁻¹⁾/3 + 2 ˣ t⁽²⁻¹⁾/2 + 0]

= 8 + t² + t

= g(t)

The antiderivative of the given function g(t)is (8t + t³/3 + t²/2 + c).

Learn more about antiderivative here: brainly.com/question/20565614

#SPJ4

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Calculate the sum of the multiples of 4 from 0 to 1000
allochka39001 [22]

Answer:

sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

Step-by-step explanation:

This problem can be solved using concept of arithmetic progression.

The sum of n term terms in arithmetic progression is given by

sum = (2a+(n-1)d)n/2

where

a is the first term

d is the common difference of arithmetic progression

_____________________________________________________

in the problem

series is multiple of 4 starting from 4 ending at 1000

so series will look like

series: 0,4,8,12,16..................1000

a is first term so

here a is 0

lets find d the common difference

common difference is given by nth term - (n-1)th term

lets take nth term as 8

so (n-1)th term = 4

Thus,

d = 8-4 = 4

d  can also be seen 4 intuitively as series is multiple of four.

_____________________________________________

let calculate value of n

we have last term as 1000

Nth term can be described

Nth term = 0+(n-1)d

1000 =   (n-1)4

=> 1000 = 4n -4

=> 1000 + 4= 4n

=> n = 1004/4 = 251

_____________________________________

now we have

n = 1000

a = 0

d = 4

so we can calculate sum of the series by using formula given above

sum = (2a+(n-1)d)n/2

       = (2*0 + (251-1)4)251/2

       = (250*4)251/2

     = 1000*251/2 = 500*251 = 125,500

Thus, sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

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What is the value of p in the linear equation 24p + 12 – 18p = 10 + 2p – 6?
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we have

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