If $F(x)=\int_{1}^{x} f(t) ~d t$ where $f(t)=\int_{1}^{t^{2}} \frac{\sqrt{2+u^{2}}}{u}$ find $F^{\prime \prime}(2)$ .
A) $3 \sqrt{2}$
B) $6 \sqrt{2}$
C) $6 \sqrt{3}$
D) $\frac{3 \sqrt{2}}{2}$
E) $3 \sqrt{3}$

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Quizplus · Accepted Answer

By the Second Fundamental Theorem of Calculus,$$F'(x)=f(x)=\int_{1}^{x^{2}} \frac{\sqrt{2+u^{2}}}{u} ~du$$and$$F''(x)=f'(x)=\frac{d}{dx} \int_{1}^{x^{2}} \frac{\sqrt{2+u^{2}}}{u} ~du=\frac{\sqrt{2+(x^{2})^{2}}}{x^{2}}=\frac{\sqrt{2+x^{4}}}{x^{2}}$$So,$$F''(2)=\frac{\sqrt{2+2^{4}}}{2^{2}}=\frac{\sqrt{2+16}}{4}=\frac{\sqrt{18}}{4}=\frac{3\sqrt{2}}{4}$$

If F(x)=∫1xf(t) dtF(x)=\int_{1}^{x} f(t) ~d tF(x)=∫1x​f(t) dt Where f(t)=∫1t22+u2uf(t)=\int_{1}^{t^{2}} \frac{\sqrt{2+u^{2}}}{u}f(t)=∫1t2​u2+u2​​ Find F′′(2)F^{\prime \prime}(2)F′′(2) A) 323 \sqrt{2}32​

If $F(x)=\int_{1}^{x} f(t) ~d t$ Where $f(t)=\int_{1}^{t^{2}} \frac{\sqrt{2+u^{2}}}{u}$ Find $F^{\prime \prime}(2)$
A) $3 \sqrt{2}$