Q. 44 - Calculus

Question

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

       $\frac{d^2}{d{x}^2}{\int }_1^{e^x} f\left(t\right)g\left(t\right)dt$

Step-by-Step Solution

Answer

Ans:    $\frac{d^2}{d{x}^2}{\int }_1^{e^x} f\left(t\right)g\left(t\right)dt =e^{x}f\left(e^x\right)g\left(e^x\right)+e^{2x}{f}^{\mathrm{\prime }}\left(e^x\right)g\left(e^x\right)+e^{2x}f\left(e^x\right)g^{\mathrm{\prime }}\left(e^x\right)$

Step 1. Given information.

given expression,

            $\frac{d^2}{d{x}^2}{\int }_1^{e^x} f\left(t\right)g\left(t\right)dt$

Step 2. The objective is to find the above derivative using the Second Fundamental Theorem of Calculus.

  Recollect that if $f$ is continuous on $[a,b]$, then

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

Therefore, 

      $\begin{array}{r}\frac{d}{dx}{\int }_1^{e^x} f\left(t\right)g\left(t\right)dt=f\left(e^x\right)g\left(e^x\right)\cdot {\left(e^x\right)}^{\mathrm{\prime }}\\ =f\left(e^x\right)g\left(e^x\right)\cdot e^x\\ =e^{x}f\left(e^x\right)g\left(e^x\right)\end{array}$

Step 3. Note that the function e x f e x g e x is continuous on [ 1 , b ] and differentiable on ( 1 , b )

So, by the Second Fundamental Theorem of calculus,

    $\begin{array}{r}\frac{d^2}{d{x}^2}{\int }_1^{e^x} f\left(t\right)g\left(t\right)dt=\frac{d}{dx}\left[\frac{d}{dx}{\int }_1^{e^x} f\left(t\right)g\left(t\right)dt\right]\\ =\frac{d}{dx}\left[e^{x}f\left(e^x\right)g\left(e^x\right)\right]\\ =\frac{d}{dx}\left(e^x\right)\cdot f\left(e^x\right)g\left(e^x\right)+e^x\cdot \frac{d}{dx}\left(f\left(e^x\right)\right)\cdot g\left(e^x\right)+e^{x}f\left(e^x\right)\cdot \frac{d}{dx}\left(g\left(e^x\right)\right)\\ =e^{x}f\left(e^x\right)g\left(e^x\right)+e^x\cdot f^{\mathrm{\prime }}\left(e^x\right)\cdot e^x\cdot g\left(e^x\right)+e^{x}f\left(e^x\right)\cdot g^{\mathrm{\prime }}\left(e^x\right)\cdot e^x\\ =e^{x}f\left(e^x\right)g\left(e^x\right)+e^{2x}{f}^{\mathrm{\prime }}\left(e^x\right)g\left(e^x\right)+e^{2x}f\left(e^x\right)g^{\mathrm{\prime }}\left(e^x\right)\end{array}$

Therefore,   $\frac{d^2}{d{x}^2}{\int }_1^{e^x} f\left(t\right)g\left(t\right)dt=e^{x}f\left(e^x\right)g\left(e^x\right)+e^{2x}{f}^{\mathrm{\prime }}\left(e^x\right)g\left(e^x\right)+e^{2x}f\left(e^x\right)g^{\mathrm{\prime }}\left(e^x\right)$