Question

Qual a integral definida de f(x)= -∫₁²(x³-4x²+3x-1)dx?

Answer 1

                               2 
-[x⁴/4-4x³/3+3x²/2-x]
                               1


=-[16/4-32/3+12/2 -2  -1/4+4/3-3/2+1]

=-[16/4-32/3+6 -2  -1/4+4/3-3/2+1]

=-[16/4-32/3+5  -1/4+4/3-3/2]

=-[15/4-28/3 +5 -6/4]

=-[9/4-28/3 +5]

=-[9/4 -13/3]

=25/12

Answer 2

Olá!

$\displaystyle -\int_1^2 x^3-4x^2+3x-1\;dx=-\left(\dfrac{x^4}{4}-\dfrac{4x^3}{3}+\dfrac{3x^2}{2}-x\right)\bigg{|}_{x=1}^{x=2} =\\ \\ \\ =-\left[\left(\dfrac{2^4}{4}-\dfrac{4\cdot 2^3}{3}+\dfrac{3\cdot 2^2}{2}-2\right)-\left(\dfrac{1^4}{4}-\dfrac{4\cdot 1^3}{3}+\dfrac{3\cdot 1^2}{2}-1\right)\right]=\\ \\ \\=-\left(4-\dfrac{32}{3}+6-2-\dfrac{1}{4}+\dfrac{4}{3}-\dfrac{3}{2}+1\right)=\\ \\ \\=-\left(9-\dfrac{128+3-16+18}{12}\right)=\dfrac{25}{12}.$



Bons estudos!