How do you use the second fundamental theorem of Calculus to find the derivative of given #int cos(sqrt( t))dt# from #[6, x]#?

Question

Charles Anctil · Accepted Answer

See the explanation below.

The second Fundamental Theorem of Calculus (a.k.a The Fundamental Theorem of Calculus, Part 2) tells us that #int_6^x f(t) dt = F(x) - F(6)# where #F# is an antiderivative of #f#. That is, where #F'(x) = f(x)# Now, #F(6)# is a constant, so its derivative is #0#. Therefore, #d/dx(F(x) - F(6)) = d/dx(F(x))-d/dx(F(6))# # = d/dx(F(x)) - 0# # = F'(x) = f(x) = cos(sqrt(x))# Note that I did not find an expression for #F(x)# because I was not asked to. If you are expected to do that, use integration by parts for #2 int underbrace(sqrtx)_u underbrace(cos(sqrtx) 1/(2sqrtx) dx)_(dv)#

Adam Adkins · Answer

undefined To find the derivative of the integral $\int_{6}^{x} \cos(\sqrt{t}) \, dt$ with respect to $x$, you can use the Second Fundamental Theorem of Calculus, which states that if $f$ is continuous on an interval containing $a$ and $F(x)$ is any antiderivative of $f(x)$ on that interval, then

$$\frac{d}{dx}\left(\int_{a}^{x} f(t) \, dt\right) = f(x).$$

In your case, the function $f(t) = \cos(\sqrt{t})$ is continuous on the interval $[6, x]$.

First, find an antiderivative of $f(t)$, denoted by $F(t)$. Let $u = \sqrt{t}$, then $du = \frac{1}{2\sqrt{t}}dt$, or $2\sqrt{t}du = dt$. Then, $F(t) = \int \cos(u) \cdot 2u \, du = 2\sin(u) + C = 2\sin(\sqrt{t}) + C$, where $C$ is the constant of integration.

Now apply the Second Fundamental Theorem of Calculus:

$$\frac{d}{dx}\left(\int_{6}^{x} \cos(\sqrt{t}) \, dt\right) = \frac{d}{dx}(F(x) - F(6)) = F'(x) = 2\sin(\sqrt{x}).$$