∫ xⁿ dx = xn + 1/(n + 1) + c where n is any constant except 0 and -1.
Apply that rule to get:
∫ 5x dx
= 5 ∫ x dx [Factor out the constant]
= 5 ∫ x1 dx [Make note of the exponent for x]
= 5x1 + 1/(1 + 1) + c
= (5/2)x2 + c
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The antiderivative of a function which is equal to 0 everywhere is a function equal to 0 everywhere.
The fundamental theorum of calculus states that a definite integral from a to b is equivalent to the antiderivative's expression of b minus the antiderivative expression of a.
-e-x + C.
You can't, unless it's an initial value problem. If f(x) is an antiderivative to g(x), then so is f(x) + c, for any c at all.
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