Answer:
We know that,
dxd(ax+b)3=3a(ax+b)2
⇒(ax+b)2=3a1dxd(ax+b)3
∴(ax+b)2=dxd(3a1(ax+b)3)
Therefore, the anti derivative of (ax+b)2 is 3a1(ax+b)3.
Find an anti derivative (or integral) of function by the method of inspection (ax + b)².
An anti derivative (or integral) of function by the method of inspection (ax + b)².
Let, f(x) = (ax + b)²
Note that, [tex] \sf \large \frac{d((ax + b) {}^{3} )}{dx} = 3a(ax + b) {}^{2} [/tex]
[tex] \sf \large \therefore (ax + b) {}^{2} = \frac{1}{3a} \: \: \frac{d((ax + b) {}^{3} )}{dx} [/tex]
[tex] \sf \large= \frac{d}{dx} \bigg ( \frac{1}{3} (ax + b {}^{3} )[/tex]
[tex] \sf \large \color{red}Hence, the \: anti - derivative \: of \: (ax + b) {}^{2} \: is \: \frac{1}{3a} (ax + b) {}^{3} .[/tex]
Hope you have satisfied. ⚘
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Answer:
We know that,
dxd(ax+b)3=3a(ax+b)2
⇒(ax+b)2=3a1dxd(ax+b)3
∴(ax+b)2=dxd(3a1(ax+b)3)
Therefore, the anti derivative of (ax+b)2 is 3a1(ax+b)3.
please mark me as brainliest
Verified answer
Question:-
Find an anti derivative (or integral) of function by the method of inspection (ax + b)².
Given:-
To Find:-
An anti derivative (or integral) of function by the method of inspection (ax + b)².
Solution:-
Let, f(x) = (ax + b)²
Note that, [tex] \sf \large \frac{d((ax + b) {}^{3} )}{dx} = 3a(ax + b) {}^{2} [/tex]
[tex] \sf \large \therefore (ax + b) {}^{2} = \frac{1}{3a} \: \: \frac{d((ax + b) {}^{3} )}{dx} [/tex]
[tex] \sf \large= \frac{d}{dx} \bigg ( \frac{1}{3} (ax + b {}^{3} )[/tex]
Answer:-
[tex] \sf \large \color{red}Hence, the \: anti - derivative \: of \: (ax + b) {}^{2} \: is \: \frac{1}{3a} (ax + b) {}^{3} .[/tex]
Hope you have satisfied. ⚘