a) Immediato. ∫ 1 2 ( x 3 − 2 x 2 ) d x = [ x 4 4 − 2 x 3 3 ] 1 2 = ( 4 − 16 3 ) − ( 1 4 − 2 3 ) = − 11 12 ≈ − 0.9167. \int_1^2(x^3-2x^2)\,dx=\left[\dfrac{x^4}{4}-\dfrac{2x^3}{3}\right]_1^2=\left(4-\dfrac{16}{3}\right)-\left(\dfrac14-\dfrac23\right)=-\dfrac{11}{12}\approx-0.9167. ∫ 1 2 ( x 3 − 2 x 2 ) d x = [ 4 x 4 − 3 2 x 3 ] 1 2 = ( 4 − 3 16 ) − ( 4 1 − 3 2 ) = − 12 11 ≈ − 0.9167.
b) Immediato. ∫ 0 π / 4 cos ( 2 x ) d x = [ sin ( 2 x ) 2 ] 0 π / 4 = sin ( π / 2 ) 2 = 1 2 . \int_0^{\pi/4}\cos(2x)\,dx=\left[\dfrac{\sin(2x)}{2}\right]_0^{\pi/4}=\dfrac{\sin(\pi/2)}{2}=\dfrac12. ∫ 0 π /4 cos ( 2 x ) d x = [ 2 sin ( 2 x ) ] 0 π /4 = 2 sin ( π /2 ) = 2 1 .
c) Per parti. Con u = 2 x u=2x u = 2 x d v = e x d x dv=e^x dx d v = e x d x 2 e x ( x − 1 ) 2e^x(x-1) 2 e x ( x − 1 ) ∫ 1 2 2 x e x d x = [ 2 e x ( x − 1 ) ] 1 2 = 2 e 2 − 0 = 2 e 2 ≈ 14.778. \int_1^2 2xe^x\,dx=\big[2e^x(x-1)\big]_1^2=2e^2-0=2e^2\approx14.778. ∫ 1 2 2 x e x d x = [ 2 e x ( x − 1 ) ] 1 2 = 2 e 2 − 0 = 2 e 2 ≈ 14.778.
d) Sostituzione u = cos x u=\cos x u = cos x ∫ tan x d x = − ln ∣ cos x ∣ \int\tan x\,dx=-\ln|\cos x| ∫ tan x d x = − ln ∣ cos x ∣ [ − ln ∣ cos x ∣ ] 0 π / 4 = − ln 2 2 = 1 2 ln 2 ≈ 0.3466. \big[-\ln|\cos x|\big]_0^{\pi/4}=-\ln\dfrac{\sqrt2}{2}=\dfrac12\ln 2\approx0.3466. [ − ln ∣ cos x ∣ ] 0 π /4 = − ln 2 2 = 2 1 ln 2 ≈ 0.3466.
e) Immediato con l’identità tan 2 x = sec 2 x − 1 \tan^2 x=\sec^2 x-1 tan 2 x = sec 2 x − 1 tan x − x \tan x-x tan x − x π / 4 < 1 \pi/4<1 π /4 < 1 [ tan x − x ] 1 π / 4 = ( 1 − π 4 ) − ( tan 1 − 1 ) = 2 − π 4 − tan 1 ≈ − 0.3428. \big[\tan x-x\big]_1^{\pi/4}=\left(1-\dfrac{\pi}{4}\right)-(\tan 1-1)=2-\dfrac{\pi}{4}-\tan 1\approx-0.3428. [ tan x − x ] 1 π /4 = ( 1 − 4 π ) − ( tan 1 − 1 ) = 2 − 4 π − tan 1 ≈ − 0.3428.
f) Immediato con sin 2 x = 1 − cos 2 x 2 \sin^2 x=\dfrac{1-\cos 2x}{2} sin 2 x = 2 1 − cos 2 x x 2 − sin 2 x 4 \dfrac{x}{2}-\dfrac{\sin 2x}{4} 2 x − 4 sin 2 x [ x 2 − sin 2 x 4 ] 1 π / 4 ≈ 0.1427 − 0.2727 = − 0.1300. \big[\tfrac{x}{2}-\tfrac{\sin 2x}{4}\big]_1^{\pi/4}\approx0.1427-0.2727=-0.1300. [ 2 x − 4 s i n 2 x ] 1 π /4 ≈ 0.1427 − 0.2727 = − 0.1300.
g) Per parti con u = arctan x u=\arctan x u = arctan x d v = d x dv=dx d v = d x ∫ arctan x d x = x arctan x − 1 2 ln ( 1 + x 2 ) \int\arctan x\,dx=x\arctan x-\dfrac12\ln(1+x^2) ∫ arctan x d x = x arctan x − 2 1 ln ( 1 + x 2 ) [ x arctan x − 1 2 ln ( 1 + x 2 ) ] 1 2 = ( 2 arctan 2 − 1 2 ln 5 ) − ( π 4 − 1 2 ln 2 ) ≈ 0.9708. \big[x\arctan x-\tfrac12\ln(1+x^2)\big]_1^2=\left(2\arctan 2-\tfrac12\ln 5\right)-\left(\tfrac{\pi}{4}-\tfrac12\ln 2\right)\approx0.9708. [ x arctan x − 2 1 ln ( 1 + x 2 ) ] 1 2 = ( 2 arctan 2 − 2 1 ln 5 ) − ( 4 π − 2 1 ln 2 ) ≈ 0.9708.
h) Divisione + frazioni parziali. Dividendo: x 4 + x x 2 − 5 x + 6 = x 2 + 5 x + 19 − 18 x − 2 + 84 x − 3 \dfrac{x^4+x}{x^2-5x+6}=x^2+5x+19-\dfrac{18}{x-2}+\dfrac{84}{x-3} x 2 − 5 x + 6 x 4 + x = x 2 + 5 x + 19 − x − 2 18 + x − 3 84 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] [ x 3 3 + 5 x 2 2 + 19 x − 18 ln ∣ x − 2 ∣ + 84 ln ∣ x − 3 ∣ ] − 1 1 = 2 3 + 38 + 18 ln 3 − 84 ln 2 ≈ 0.2156. \left[\tfrac{x^3}{3}+\tfrac{5x^2}{2}+19x-18\ln|x-2|+84\ln|x-3|\right]_{-1}^1=\tfrac23+38+18\ln 3-84\ln 2\approx0.2156. [ 3 x 3 + 2 5 x 2 + 19 x − 18 ln ∣ x − 2∣ + 84 ln ∣ x − 3∣ ] − 1 1 = 3 2 + 38 + 18 ln 3 − 84 ln 2 ≈ 0.2156.
i) Denominatore quadrato. 3 x 2 − 6 x + 3 = 3 ( x − 1 ) 2 3x^2-6x+3=3(x-1)^2 3 x 2 − 6 x + 3 = 3 ( x − 1 ) 2 2 x + 4 3 ( x − 1 ) 2 = 1 3 ( 2 x − 1 + 6 ( x − 1 ) 2 ) \dfrac{2x+4}{3(x-1)^2}=\dfrac{1}{3}\!\left(\dfrac{2}{x-1}+\dfrac{6}{(x-1)^2}\right) 3 ( x − 1 ) 2 2 x + 4 = 3 1 ( x − 1 2 + ( x − 1 ) 2 6 ) [ 4 , 5 ] [4,5] [ 4 , 5 ] 1 3 ( 2 ln 4 3 + 1 2 ) ≈ 0.3585. \tfrac13\!\left(2\ln\tfrac43+\tfrac12\right)\approx0.3585. 3 1 ( 2 ln 3 4 + 2 1 ) ≈ 0.3585.
j) Sostituzione u = x 2 u=x^2 u = x 2 d u = 2 x d x du=2x\,dx d u = 2 x d x ∫ 0 1 x e − x 2 d x = 1 2 [ − e − u ] 0 1 = 1 2 ( 1 − e − 1 ) ≈ 0.3161. \int_0^1 xe^{-x^2}dx=\tfrac12\big[-e^{-u}\big]_0^1=\tfrac12(1-e^{-1})\approx0.3161. ∫ 0 1 x e − x 2 d x = 2 1 [ − e − u ] 0 1 = 2 1 ( 1 − e − 1 ) ≈ 0.3161.
k) Sostituzione u = 1 + t 2 u=1+t^2 u = 1 + t 2 ∫ 0 1 t 1 + t 2 d t = 1 2 [ ln ( 1 + t 2 ) ] 0 1 = 1 2 ln 2 ≈ 0.3466. \int_0^1\dfrac{t}{1+t^2}dt=\tfrac12\big[\ln(1+t^2)\big]_0^1=\tfrac12\ln 2\approx0.3466. ∫ 0 1 1 + t 2 t d t = 2 1 [ ln ( 1 + t 2 ) ] 0 1 = 2 1 ln 2 ≈ 0.3466.
l) Per parti (u = ln t u=\ln t u = ln t d v = d t dv=dt d v = d t ∫ ln t d t = t ln t − t \int\ln t\,dt=t\ln t-t ∫ ln t d t = t ln t − t [ t ln t − t ] 1 2 = ( 2 ln 2 − 2 ) − ( − 1 ) = 2 ln 2 − 1 ≈ 0.3863. \big[t\ln t-t\big]_1^2=(2\ln 2-2)-(-1)=2\ln 2-1\approx0.3863. [ t ln t − t ] 1 2 = ( 2 ln 2 − 2 ) − ( − 1 ) = 2 ln 2 − 1 ≈ 0.3863.
m) Per parti (due volte): ∫ e x cos x d x = e x ( sin x + cos x ) 2 \int e^x\cos x\,dx=\dfrac{e^x(\sin x+\cos x)}{2} ∫ e x cos x d x = 2 e x ( sin x + cos x ) [ e x ( sin x + cos x ) 2 ] 0 π / 2 = e π / 2 ⋅ 1 2 − 1 2 = e π / 2 − 1 2 ≈ 1.9052. \left[\dfrac{e^x(\sin x+\cos x)}{2}\right]_0^{\pi/2}=\dfrac{e^{\pi/2}\cdot 1}{2}-\dfrac{1}{2}=\dfrac{e^{\pi/2}-1}{2}\approx1.9052. [ 2 e x ( sin x + cos x ) ] 0 π /2 = 2 e π /2 ⋅ 1 − 2 1 = 2 e π /2 − 1 ≈ 1.9052.
n) Immediato. ∫ 0 1 x d x = [ 2 3 x 3 / 2 ] 0 1 = 2 3 ≈ 0.6667. \int_0^1\sqrt{x}\,dx=\left[\tfrac23 x^{3/2}\right]_0^1=\dfrac23\approx0.6667. ∫ 0 1 x d x = [ 3 2 x 3/2 ] 0 1 = 3 2 ≈ 0.6667.
o) Immediato. x x = x 3 / 2 x\sqrt{x}=x^{3/2} x x = x 3/2 ∫ 0 1 x 3 / 2 d x = [ 2 5 x 5 / 2 ] 0 1 = 2 5 = 0.4. \int_0^1 x^{3/2}dx=\left[\tfrac25 x^{5/2}\right]_0^1=\dfrac25=0.4. ∫ 0 1 x 3/2 d x = [ 5 2 x 5/2 ] 0 1 = 5 2 = 0.4.
a = − 11 12 , b = 1 2 , c = 2 e 2 , d = 1 2 ln 2 , … , n = 2 3 , o = 2 5 \boxed{a=-\tfrac{11}{12},\ b=\tfrac12,\ c=2e^2,\ d=\tfrac12\ln2,\ \dots,\ n=\tfrac23,\ o=\tfrac25} a = − 12 11 , b = 2 1 , c = 2 e 2 , d = 2 1 ln 2 , … , n = 3 2 , o = 5 2