Question

Com base na Integral por partes, determine f(x) = 2x lnx no intervalo (1, 2).

Answer 1

2 ∫ x * ln(x) dx

u =ln x ==> du=1/x dx

x dx = dv .... ∫ x dx = ∫ dv ==>x²/2 =v

∫ x * ln(x) dx = ln x * x²/2 - ∫ (x²/2)* 1/x dx

∫ x * ln(x) dx = ln x * x²/2 - (1/2)*∫ x dx

∫ x * ln(x) dx = ln x * x²/2 - x²/4

2 ∫ x * ln(x) dx = ln x*x² -x²/2 = (x²/2)* (2*ln(x) -1)

2

[(x²/2)* (2*ln(x) -1)] ≈

1

=(2 * (2*ln(2) -1) - (0,5 * (2*ln(1) -1)

=4 ln(2) - 2 +0,5

=2,7726 -2 +0,5 ≈ 1,2726

Answer 2

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            $\Huge\boxed{\boxed{\boxed{\boxed{\sf Integral~\rm por~\tt partes}}}}$

$\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int u\cdot dv=u\cdot v-\int v\cdot du}}}}$

$\displaystyle\sf\int 2x\ell nx~dx\\\underline{\rm fac_{\!\!,}a}~u=\ell nx\implies du=\dfrac{1}{x}dx\\\sf dv=2x~dx\implies v=x^2\\\displaystyle\sf\int 2x\ell nx~dx=\ell nx\cdot x^2-\int\diagup\!\!\!\!\!x^2\cdot\dfrac{1}{\diagup\!\!\!x}~dx\\\displaystyle\sf\int 2x\ell nx~dx=x^2\ell nx-\int x~dx\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int 2x\ell nx~dx=x^2\ell nx-\dfrac{1}{2}x^2+k}}}}\blue{\checkmark}$

$\displaystyle\sf\int_1^2 2x\ell nx~dx=\bigg[x^2\ell nx-\dfrac{1}{2}x^2\bigg]_1^2\\\sf =2^2\ell n(2)-\dfrac{1}{2}\cdot2^2-\bigg[1^2\ell n1-\dfrac{1}{2}\cdot1^2\bigg]\\\sf 4\ell n(2)-2-0+\dfrac{1}{2}=\dfrac{8\ell n(2)-4-0+1}{2}\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int_1^22x\ell n(x)~dx=\dfrac{8\ell n(2)-3}{2}}}}}\blue{\checkmark}$