Question

Use the Second Fundamental Theorem of Calculus, if needed, to calculate each of the derivatives given below.

$\frac{d^2}{d{x}^2}{\int }_2^{3x} \left(t^2+1\right)dt$

Short answer

Ans: $\frac{d^2}{d{x}^2}{\int }_2^{3x} \left(t^2+1\right)dt=54x$

Step-by-step solution

Step 1. Given information.

given expression,

$\frac{d^2}{d{x}^2}{\int }_2^{3x} \left(t^2+1\right)dt$

Step 2.  The objective is to calculate the derivative.

Now, if$f$is continuous on$[a,b]$then for all$x\in [a,b]$,

$\frac{d}{dx}{\int }_a^{u\left(x\right)} f\left(t\right)dt=f\left(u\right(x\left)\right)u^{\mathrm{\prime }}\left(x\right)$

So,

$\begin{array}{r}f\left(u\right(t\left)\right)=\left((3t{)}^2+1\right)\\ f\left(u\right(x\left)\right)=\left((3x{)}^2+1\right)\\ u^{\mathrm{\prime }}\left(x\right)=3\\ f\left(u\right(x\left)\right)u\left(x\right)=3\left(3x^2+1\right)\end{array}$

Step 3. The derivate expression can be written as,

$\begin{array}{r}\frac{d^2}{d{x}^2}{\int }_2^{3x} \left(t^2+1\right)dt\\ =\frac{d}{dx}\left(\frac{d}{dx}{\int }_2^{3x} \left(t^2+1\right)dt\right)\\ =\frac{d}{dx}\left(3\left((3x{)}^2+1\right)\right)\\ =3\frac{d}{dx}\left((3x{)}^2+1\right)\\ =3\frac{d}{dx}\left(9x^2+1\right)\\ =3\left(18x\right)+0\\ =54x\end{array}$

Therefore, the answer is$54x$