Definite Integrals = Fundamental Theorem of Calculus Part 2

Definition: Fundamental Theorem of Calculus Part 2 = Definite Integral

If you have a continuous function, $f\left(x\right)$, on a closed x-interval [ a , b ], then the result of the antiderivative will be $F\left(b\right)–F\left(a\right)$.

$\underset{a}{\overset{b}{\int }}f\left(x\right)\mathit{dx}\mathit{=}{F\left(x\right)|}_a^b\mathit{=}F\left(b\right)–F\left(a\right)\mathit{=}$ exact net area between a curve and the x-axis .

What the notation is directing you to do is to; 1) take the antiderivative ( undo the derivative) of your given equation, and then 2) evaluate by plugging the upper limit , b ,into the result and subtracting the lower limit , a ,plugged into the result.

I always say Top – Bottom . Plug the top value in and subtract the bottom value plugged in. Whenever in doubt, math rules generally work under that top minus bottom rule.

The answer to a Definite Integral represents the Net Area between the curve $f\left(x\right)$ and the x-axis on a closed x-interval [ a , b ], a definite value.

Note: Since the Fundamental Theorem of Calculus Part 2 has a specific x-interval [ a , b ], this means that the lower bound, a ,will always be the smaller x-value ,and the upper bound, b ,will always be the larger x-value .

If your integral’s bounds are not setup with the lower bound as the smaller x-value and the upper bound as the larger x-value , then you will have to flip the bounds by performing a rewrite on the definite integral. To flip the bounds, you put a negative sign (–) in front of the integral.

$\underset{a}{\overset{b}{\int }}f\left(x\right)\mathit{dx}\mathit{=}\mathit{–}\underset{b}{\overset{a}{\int }}f\left(x\right)\mathit{dx}$

$\underset{5}{\overset{3}{\int }}f\left(x\right)\mathit{dx}\mathit{=}\mathit{–}\underset{3}{\overset{5}{\int }}f\left(x\right)\mathit{dx}$