§5.2 The Second Fundamental Theorem of Calculus (continued)

Differentiating an Integral Function

Summary

Compute $\frac{d}{dx}\left[\int_4^x e^{t^2}\,dt\right]$.

Compute $\frac{d}{dx}\left[\int_x^1\cos(t^3)\,dt\right]$.

Compute $\frac{d}{dx}\left[\int_4^{x^3}\sin(t^2)\,dt\right]$.

For $F(x) = \int_{2}^x g(t)\,dt$ with shaded areas $A_2 = 12.75$ total above the axis on $[0.5,4]$ of which $6.06$ lies on $[0.5,2]$, the value $F(4)$ equals

An aircraft's climb rate $c(h)$ is measured in feet per minute. The function $m(h)$ giving minutes needed per additional foot at altitude $h$ is

For $\int_{-2}^{x}\frac{d}{dt}\left[\frac{t^4}{1+t^4}\right]dt$, the value equals $\frac{x^4}{1+x^4} - \frac{16}{17}$. Evaluate this at $x = 0$, rounded to three decimals.

For $\int_3^x\frac{d}{dt}[\ln(1+t^2)]\,dt = \ln(1+x^2) - \ln(10)$, evaluate at $x = 1$, rounded to three decimals.

With $F(4) = 6.69$ and a further signed area of $-0.36$ on $[4,5]$, compute $F(5)$.

Using midpoint samples $m(1000) + m(3000) + m(5000)$ each times $2000$ feet, the climb time to $6000$ feet is approximately $\left(\frac{1}{875} + \frac{1}{780} + \frac{1}{685}\right) \cdot 2000$ minutes. Compute this value, rounded to two decimals.

If sand starts at $2500$ cubic meters with pumping rate $S(t)$ and tide removal rate $R(t)$, then $Y(x) = 2500 + \int_0^x[S(t)-R(t)]\,dt$ models the sand present at time $x$.