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Question Number 58639 by rahul 19 last updated on 26/Apr/19

Answered by tanmay last updated on 27/Apr/19

$$F\left(x\right)={\int }_0^{x}tf\left(t\right)dt$$$$\frac{dF}{dx}={\int }_0^x\frac{\partial }{\partial x}(tf\left(t\right)dt+xf\left(x\right)\frac{dx}{dx}-0\times f\left(0\right)\times \frac{d0}{dx}$$$$\frac{dF}{dx}=xf\left(x\right)$$$$F\left(x^2\right)=x^4+x^5=x^4(1+x)={\left(x^2\right)}^2(1+\sqrt{x^2})$$$$F\left(x\right)=x^2(1+\sqrt{x})=x^2+x^{\frac{5}{2}}$$$$\frac{dF}{dx}=2x+\frac{5}{2}{x}^{\frac{3}{2}}$$$$2x+\frac{5}{2}{x}^{\frac{3}{2}}=xf\left(x\right)$$$$f\left(x\right)=2+\frac{5}{2}\sqrt{x}$$$$a)f\left(4\right)=2+\frac{5}{2}\times \sqrt{4}=7$$$$b)f\left(x\right)=2+\frac{5}{2}\sqrt{x}continousfunction[x>0]$$$$c)f\left(x\right)=2+\frac{5}{2}\sqrt{x}$$$$\frac{df}{dx}=\frac{5}{2}\times \frac{1}{2\sqrt{x}}=\frac{5}{4\sqrt{x}}whenx>0\frac{df}{dx}>0soincreasinggunction$$

Commented by rahul 19 last updated on 27/Apr/19

$$thankssir!$$

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