Integrals

${\displaystyle \underset{0}{\overset{1}{\int }}\left[-8x^2+6x-1\right]dx}$  

Let f (x) = -8x² + 6x – 1

f (0) = -1, f (1) = -3

max f (x) = $-\frac{D}{4a}=-\frac{36-32}{-32}=\frac{1}{8}$  

= $-\frac{1}{4}-1\left(\frac{3}{4}-\frac{\lambda }{2}\right)-2\left(\frac{\sqrt[]{17}-3}{8}-\frac{3}{4}\right)-3\left(1-\frac{\sqrt[]{17}-3}{8}\right)$  

= $\frac{\sqrt[]{17}-13}{8}$

 ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{3+2sinx+cosx}Puttan\frac{x}{2}=t\Rightarrow \frac{1}{2}se{c}^2\frac{x}{2}dx=dt}$

$={\displaystyle \underset{0}{\overset{1}{\int }}\frac{2dt}{2t^2+4t+4}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{{\left(t+1\right)}^2+1}={\left[ta{n}^{-1}\left(t+1\right)\right]}_0^1}=ta{n}^{-1}2-ta{n}^{-1}1}$

 $\mathrm{t}\mathrm{a}\mathrm{n}?\theta =\frac{10}{\mathrm{x}}=\frac{\mathrm{h}}{{\mathrm{x}}_2}\Rightarrow {\mathrm{x}}_2=\frac{\mathrm{h}\mathrm{x}}{10}$

$\mathrm{t}\mathrm{a}\mathrm{n}??=\frac{15}{\mathrm{x}}=\frac{\mathrm{h}}{\mathrm{x}}\Rightarrow {\mathrm{x}}_1=\frac{\mathrm{h}\mathrm{x}}{15}$

Now,  $x_1+x_2=x=\frac{hx}{15}+\frac{hx}{10}$

$\Rightarrow 1=\frac{\mathrm{h}}{10}+\frac{\mathrm{h}}{15}\Rightarrow \mathrm{h}=6$