Let f be f(x) = xⁿ its antiderivative is F(x) = 1/n+1x^n+1+C (Read F(x) equals 1 over n+1 times x exponent n+1 plus the constant of integration C) . You can rewrite the expression by utilizing a horizontal bar for the fraction 1 over n+1. You can also rewrite the expression x^n+1 by putting n+1 besides x and above. The rewriting will make the expression clearer.
The antiderivative of f(x) being the indefinite integral, we can write ∫f(x)dx = 1/n+1.x^n+1+C (Read sum f(x) dx equals 1 over n+1 times x exponent n+1 plus the constant of integration C).
In order to prove this, we have to demonstrate that the derivative of 1/n+1.x^n+1+C is xⁿ.
Let's calculate the derivative of F(x) = 1/n+1.x^n+1+C
F'(x) = n+1/n+1x^n+1-1 (Read n+1 over n+1 times x exponent n+1-1. By simplifying and reducing F'(x) = xⁿ
Integral of the inverse function
The integral of the inverse function f(x) = 1/x is ∫1/xdx = lnx +C.
Proof
The derivative of F(x) = lnx is F'(x) = 1/x
Integral of the exponential function
The integral of the exponential function f(x) = e^x is ∫e^xdx = e^x + C. The term e^x reads e exponential x.
Proof
The derivative of F(x) = e^x is F'(x) = e^x.
Rules of integrable functions
If f and g are integrable functions and C is a constant, then
∫[f(x)+g(x)] = ∫f(x)dx+∫g(x)dx
∫[f(x)-g(x)] = ∫f(x)dx-∫g(x)dx
∫kf(x)dx = k∫f(x)dx
Example
Solution
Using the basic rules of integration we have:
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