Question

f'(x)= 12x⁵+30x² and f(4)= -23​
Please give answer in steps
This is from integration

Answer 1

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

Given that

$\rm :\longmapsto\:f'(x) = {12x}^{5} + {30x}^{2}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{d}{dx} f(x) = {12x}^{5} + {30x}^{2}$

On integrating both sides, we get

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{d}{dx} f(x) \: dx=\displaystyle\int\sf ({12x}^{5} + {30x}^{2}) \: dx$

$\rm :\longmapsto\:f(x) \: =\displaystyle\int\sf {12x}^{5}dx + \displaystyle\int\sf {30x}^{2} \: dx$

$\rm :\longmapsto\:f(x) \: =12\displaystyle\int\sf {x}^{5}dx +30 \displaystyle\int\sf {x}^{2} \: dx$

We know that

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

So, using this

$\rm :\longmapsto\:f(x) = 12 \times \dfrac{ {x}^{5 + 1} }{5 + 1} + 30 \times \dfrac{ {x}^{2 + 1} }{2 + 1} + k$

$\rm :\longmapsto\:f(x) = 12 \times \dfrac{ {x}^{6} }{6} + 30 \times \dfrac{ {x}^{3} }{3} + k$

$\rm :\longmapsto\:f(x) = {2x}^{6} + {10x}^{3} + k$

Now, given that

$\rm :\longmapsto\:f(4) = - 23$

$\rm :\longmapsto\:{2(4)}^{6} + {10(4)}^{3} + k = - 23$

$\rm :\longmapsto\:8192 + 640 + k = - 23$

$\rm :\longmapsto\:8832 + k = - 23$

$\rm :\longmapsto\:k = - 23 - 8832$

$\bf :\longmapsto\:k = - 8855$

Thus,

$\bf :\longmapsto\:f(x) = {2x}^{6} + {10x}^{3} - 8855$

Additional Information :-

$\boxed{ \sf{ \:\displaystyle\int\sf k \: dx = kx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\sf sinx \: dx = - cosx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\sf cosx \: dx = sinx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\sf cotx \: dx = log \: sinx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\sf tanx \: dx = log \: secx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\sf secx \: dx = log \: (secx + tanx) + c}}$

$\boxed{ \sf{ \:\displaystyle\int\sf cosecx \: dx = log \: (cosecx - cotx) + c}}$

$\boxed{ \sf{ \:\displaystyle\int\sf {sec}^{2}x \: dx = tanx + c}}$

$\boxed{ \sf{ \:\displaystyle\int\sf {cosec}^{2}x \: dx = - \: cotx + c}}$