"You are standing on a cliff, ready to jump into the ocean. The ocean is 30 feet below you. What is your velocity as you hit the water?". I know that I can solve it with $Δx=0.5at^2$, and use $t*a$ to get the final velocity.
However, I want to know if there is a way to solve it with integral. First, start with finding $t$. $$\int_0^x 32.174t\,dt= 30$$ With the formula, I know that $x=1.3656sec$, but how do I solve the upper bound for the integral to get that answer?
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In general, when solving integrals you get the antiderivative function $F(X) = \int{f(x)dx}$, but in order to eliminate to possibility of infinite number in constants $C$ added after the integration we can use one of two approaches:
Definite integration in an interval $[a,b]$ when we know both $a$ and $b$: $$\int_a^b{f(x)dx}=[F(x)+C]_a^b= F(b)+c-(F(a)+C) = F(b)-F(a)$$
Using another varaible as upper bound in order to get a function as a result: for example $$\ln(x) = \int_1^x{\frac{1}{t}dt}= \ln(x)+C-({\ln\not{1}}+C)$$ this is usfull in times where the lower bound gives us a zero (either by having lower bound equal $0$ like your problem, or $1$ in the natural logarithm that will turn into $0$ after the integration.
Both are eligible and legitimate, but sometimes the latter is more relevant since the problem does not imply specific bounds $[a,b]$ for the interval.
I assume like the following
$$ \int_{0}^{x} 32.174 t \textrm{d}t = 30 \\ \int_{0}^{x} t \textrm{d}t = \frac{30} {32.174} \\ \frac{t^{2}}{2} \Big|_0^x = \frac{30}{32.174} \\ \frac{x^{2}}{2} = \frac{30}{32.174} \\ x = \sqrt{\frac{60}{32.174}}$$