2 variable Integration

[jacks]

If $f(x)$ is a Continuous function and $\displaystyle F(x) = \int_{0}^{x}\left((2t+3)\cdot \int_{1}^{2}f(u)du\right)dt\;,$ Then $|F''(x)|\cdot |f(2)| = $

[Papapetros Vaggelis]

Hi jacks.

Since the function \(\displaystyle{f}\) is continuous ( i suppose that you mean at \(\displaystyle{\mathbb{R}}\)) , we have that :

\(\displaystyle{\begin{aligned} F(x)&=\int_{0}^{x}\left((2\,t+3)\,\int_{1}^{2}f(u)\,\mathrm{d}u\right)\,\mathrm{d}t\\&=\int_{1}^{2}f(u)\,\mathrm{d}u\,\int_{0}^{x}\left(2\,t+3\right)\,\mathrm{d}t\\&=\int_{1}^{2}f(u)\,\mathrm{d}u\,\left[t^2+3\,t\right]_{0}^{x}\\&=\int_{1}^{2}f(u)\,\mathrm{d}u\,\left(x^2+3\,x\right)\,\,,x\in\mathbb{R}\end{aligned}}\)

Now, \(\displaystyle{F^\prime(x)=\left(2\,x+3\right)\,\int_{1}^{2}f(u)\,\mathrm{d}u\,\,,x\in\mathbb{R}}\) and

\(\displaystyle{F^{\prime \prime}(x)=2\,\int_{1}^{2}f(u)\,\mathrm{d}u\,,x\in\mathbb{R}}\) .

Therefore,

\(\displaystyle{\left|F^{\prime \prime}(x)\right|\,\left|f(2)\right|=\left|2\,f(2)\,\int_{1}^{2}f(u)\,\mathrm{d}u\right|}\) .