Question

Evaluate the following integrals ...​

Answer 1

Answer:

Please mark BRAINLIEST...

Answer 2

$\large\underline{\bold{Given \:Question - }}$

$\sf \: f'(x) = {6x}^{2} - 4x + 5, \: f(0) = 5, \: then \: find \: f(x).$

$\red{\large\underline{\sf{Solution-}}}$

Given function is

$\rm :\longmapsto\: \sf \: f'(x) = {6x}^{2} - 4x + 5$

On integrating both sides w. r. t. x, we get

$\rm :\longmapsto\: \sf \: \displaystyle\int\sf f'(x) \: dx= \displaystyle\int\sf \bigg[ {6x}^{2} - 4x + 5\bigg] \: dx$

$\rm :\longmapsto\: \sf \: f(x)= \displaystyle\int\sf {6x}^{2}dx - \displaystyle\int\sf 4xdx + \displaystyle\int\sf 5 \: dx$

$\rm :\longmapsto\: \sf \: f(x) =6 \displaystyle\int\sf {x}^{2}dx -4 \displaystyle\int\sf xdx +5 \displaystyle\int\sf \: dx$

We know,

$\boxed{ \bf{ \:\displaystyle\int\sf {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1} + c \: \: }}$

So, using this, we get

$\sf :\longmapsto\:f(x) = 6\bigg[\dfrac{ {x}^{3} }{3} \bigg] - 4\bigg[\dfrac{ {x}^{2} }{2} \bigg] + 5x + c$

$\sf :\longmapsto\:f(x) = {2x}^{3} - {2x}^{2} + 5x + c$

On substituting x = 0, we get

$\sf :\longmapsto\:f(0) = {2(0)}^{3} - {2(0)}^{2} + 5(0) + c$

$\bf\implies \:c = 5$

Thus,

$\sf :\longmapsto\:\boxed{ \bf{ \: \: \:f(x) = {2x}^{3} - {2x}^{2} + 5x + 5 \: \: \: }}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$