3 Theorem If f is continuous on [a, b], or if f has only a finite number discontinuities, then f is integrable on [a, b]; that is, the definite integral exists. If f is integrable on [a, b], then the limit in Definition 2 exists and give value no matter how we choose the sample points x. To simplify the calcul integral we often take the sample points to be right endpoints. Then x = x,a nition of an integral simplifies as follows. 4 Theorem If f is integrable on [a, b], then Cod - |f(x) dx = lim E f(x) Ax i-1 where Ax = and Xi = a + i Ax EXAMPLE 1 Express lim (x + x, sin x,) Ax as an integral on the interval [0, 7]. SOLUTION Comparing the given limit with the limit in Theorem 4, we see t will be identical if we choose f(x) = x' + x sin x. We are given that a Therefore, by Theorem 4, we have %3D L² + "(x' (x+x sin x) dx lim (x + x, sin x,) Ax = 1-1 Lotar when we apply the definite integral to physical situations, it will be in did in Example 1. When Leibni eWe kaas that Si fde = Ax (atiax) flati Ax) %3D 下=L fC)こxtx b-a_of2) %3D धहे 2. (2+i-) lim (-2ti- lim T-1

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Use the form of the definition of the integral given in Theorem 4 (see image) to evaluate the integral. 

 

I have included a picture of my work so far as well, and in the steps to solving this problem, can you especially please show me how we get from step 1 to step 2, where we square the (-2+(2i)/n)?

3 Theorem If f is continuous on [a, b], or if f has only a finite number
discontinuities, then f is integrable on [a, b]; that is, the definite integral
exists.
If f is integrable on [a, b], then the limit in Definition 2 exists and give
value no matter how we choose the sample points x. To simplify the calcul
integral we often take the sample points to be right endpoints. Then x = x,a
nition of an integral simplifies as follows.
4 Theorem If f is integrable on [a, b], then
Cod -
|f(x) dx = lim E f(x) Ax
i-1
where
Ax =
and
Xi = a + i Ax
EXAMPLE 1 Express
lim (x + x, sin x,) Ax
as an integral on the interval [0, 7].
SOLUTION Comparing the given limit with the limit in Theorem 4, we see t
will be identical if we choose f(x) = x' + x sin x. We are given that a
Therefore, by Theorem 4, we have
%3D
L² +
"(x'
(x+x sin x) dx
lim (x + x, sin x,) Ax =
1-1
Lotar when we apply the definite integral to physical situations, it will be in
did in Example 1. When Leibni
Transcribed Image Text:3 Theorem If f is continuous on [a, b], or if f has only a finite number discontinuities, then f is integrable on [a, b]; that is, the definite integral exists. If f is integrable on [a, b], then the limit in Definition 2 exists and give value no matter how we choose the sample points x. To simplify the calcul integral we often take the sample points to be right endpoints. Then x = x,a nition of an integral simplifies as follows. 4 Theorem If f is integrable on [a, b], then Cod - |f(x) dx = lim E f(x) Ax i-1 where Ax = and Xi = a + i Ax EXAMPLE 1 Express lim (x + x, sin x,) Ax as an integral on the interval [0, 7]. SOLUTION Comparing the given limit with the limit in Theorem 4, we see t will be identical if we choose f(x) = x' + x sin x. We are given that a Therefore, by Theorem 4, we have %3D L² + "(x' (x+x sin x) dx lim (x + x, sin x,) Ax = 1-1 Lotar when we apply the definite integral to physical situations, it will be in did in Example 1. When Leibni
eWe kaas that Si fde = Ax (atiax)
flati Ax)
%3D
下=L
fC)こxtx
b-a_of2)
%3D
धहे
2.
(2+i-)
lim
(-2ti-
lim
T-1
Transcribed Image Text:eWe kaas that Si fde = Ax (atiax) flati Ax) %3D 下=L fC)こxtx b-a_of2) %3D धहे 2. (2+i-) lim (-2ti- lim T-1
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