Help me Solve the following Definite Integral in detail: $$\int\limits_{1}^{4}\frac{ 3x^3-2x^2+4 }{ x^2}$$

Question

anonymous · Answer

$$
\int\limits_{1}^{4}\frac{ 3x^3-2x^2+4 }{ x^2}dx=\
\int\limits_{1}^{4} \left( 3 x -2 - \frac 4 {x^2}     \right)dx

$$
Can you finish it now?

anonymous · Answer

I can not  :(

anonymous · Answer

$$

\left[     
\frac {3 x^2}2 - 2 x - \frac 4 x  \right]_1^4
$$

anonymous · Answer

My first post has a sign misprint. Let me correct ti now$$
\int\limits_{1}^{4}\frac{ 3x^3-2x^2+4 }{ x^2}dx=\
\int\limits_{1}^{4} \left( 3 x -2 + \frac 4 {x^2}     \right)dx$$

anonymous · Answer

ok

anonymous · Answer

the result is this last part that you posted?

anonymous · Answer

No, you have to compute the value of the expression at 4 than at 1 and you subtract the two values

anonymous · Answer

If we call
$$
F(x)=\frac {3 x^2}2 - 2 x - \frac 4 x\
$$
The answer is
$$
F(4)- F(1)
$$

anonymous · Answer

$$
F(4)=15\
F(1)=-\frac 92\
F(4)-F(1)=\frac {39}2
$$

anonymous · Answer

= (3/2 . 4² - 2.4 - 4/4) - (3/2.1² - 2.1 - 4/1) 
=(24 - 8 - 1) - ( 3/2 - 6) 
= 15 - (-9/2) 
= 15 + 9/2 
= 39/2

anonymous · Answer

Yes

anonymous · Answer

Thank you friend ..

anonymous · Answer

YW