=lim(x→∞)e^[((x- 1)/2)ln((x+3)/(x+6))]

=e^[ 1/2？ lim(x→∞)(x- 1)ln((x+3)/(x+6))]

∫lim(x→∞)(x- 1)ln((x+3)/(x+6))

= lim(x→∞)ln((x+3)/(x+6))/[ 1/(x- 1)]

type 0/0(lim(x→∞)ln((x+3)/(x+6))= lim(x→∞)ln( 1-3/(x+6))= ln 1 = 0)。 For L'H?pital, the upper and lower derivatives are simultaneous.

=lim(x→∞) ((x+6)/(x+3))？ ((x+3)/(x+6))'/[ 1/(x- 1)]'

PS: Pay attention to the derivation of (ln ((x+3)/(x+6))' composite function.

= lim(x→∞) 1/((x+3)(x+6))/[- 1/(x- 1)^2]

=lim(x→∞)(x- 1)^2/((x+3)(x+6))

= lim (x →∞) (x2-2x+1)/(x2+9x+18) See the coefficient ratio before the highest term for the rational fraction. The denominator contains the highest term, and the coefficient before x 2 is 1. The score of 1 contains the coefficients before x 2.

= 1

∴ Original formula = e (1/2? 1)=e^ 1/2

2.∫(0→ 1)(x^2+x-2)e^xdx

=∫(0→ 1)x？ dx+∫(0→ 1)xdx-2∫(0→ 1)e^xdx

=[(x？ e^x)|(0→ 1)-∫(0→ 1)e^xdx？ ]+∫(0→ 1)xdx-2∫(0→ 1)e^xdx

=[(x？ e^x)|(0→ 1)-2∫(0→ 1)xe^xdx]+∫(0→ 1)xdx-2e^x|(0→ 1)

=(x？ e^x)|(0→ 1)-∫(0→ 1)xe^xdx -2e^x|(0→ 1)

=(x？ e^x)|(0→ 1)-∫(0→ 1)xde^x -2e^x|(0→ 1)

=(x？ e^x)|(0→ 1)-[xe^x |(0→ 1)-∫(0→ 1)e^xdx]-2e^x|(0→ 1)

=(x？ e^x)|(0→ 1)-xe^x |(0→ 1)+e^x|(0→ 1)-2e^x|(0→ 1)

=(x？ e^x)|(0→ 1)-xe^x |(0→ 1)-e^x|(0→ 1)

=2e-2

3. according to the meaning of the question dA=x? Advanced (short for deluxe)

A=∫(a→(a+ 1))x？ Advanced (short for deluxe)

=x？ /3|(a→(a+ 1))

=((a+ 1)？ -a？ )/3

=((a+ 1)-a)((a+ 1)？ +a(a+ 1)+a？ )/3

=(3a？ +3a+ 1)/3

=a？ +a+ 1/3

=(a+ 1/2)？ + 1/ 12

When a=- 1/2. Amin =112

At this time dV=∏(x? )? dx=∏x^4dx

v=∫(- 1/2→ 1/2)∏x^4dx

=∏/5 ? x^5|(- 1/2→ 1/2)

=∏/80