Question

Evaluate the integral

Answer 1

\[Integral (x ^2 +e ^{x})\]

Answer 2

Is it a defined or undefined integral?

Answer 3

Undefined

Answer 4

I won't prove it, but I'll give you the formula, say we have:
\[\int\limits_{}^{}f(x)+g(x) dx\]
When we have the integral of the sum or sustraction of two or more functions, it's equal to the independent integral of all the terms, meaning that:
\[\int\limits_{}^{}f(x)+g(x) dx=\int\limits f(x) dx + \int\limits g(x)dx\]

Answer 5

if f(x) is continous and if G (x)=f(x) then

Answer 6

\[\int\limits_{a}^{b} f(x) dx=G(B)-g(A)\]

Answer 7

That's what we call "barrows rule".
So, given f(x) continous in the interval [a,b] and let G(x) be any primitive of f, so that means G'(x)=f(x).
Then:
\[\int\limits_{a}^{b}=f(x)dx=G(b)-G(a)\]
But let's state my hypothesis for this theorem, shall we?:
H) f continous in [a,b]
U'(x)=f(x)

\[T) \int\limits_{a}^{b}f(x)dx=G(b)-G(a)\]
Proof:
by hypothesis:
\[U(x)=\int\limits_{a}^{x}f(x)dx\]
So by the first fundamental theorem of calculus, we got:
\[U'(x)=f(x)=G'(x), \forall x \in [a,b]\]
Therefore:
\[\exists c \in \mathbb{R} / \forall x \in [a,b] , U(x)=G(x)+c\]
We can observe the following (becuse I substituted x for a):
\[U(a)=G(a)+c=0\]
in other for that to be true, "c" must be equal to -U(a):
\[U(x)=G(x)-G(a)=0\]
Since we can give "x" any value we wish, let's make x=b, and by the initial integral I stated on the beginning:
\[if: U(x)=\int\limits_{a}^{x}f(x)dx\]
\[U(x)=G(x)-G(a)\]
\[x=b\]
Therefore:
\[\int\limits_{a}^{b} f(x) dx = G(b)-G(a)\]
QED.