The value of the definite integral ∫ limits-∞ ln 3 e x dx equals [Note: y denotes fractional part of y.]

Question

The value of the definite integral ∫ limits-∞ ln 3 e x dx equals [Note: y denotes fractional part of y.] (A) 3+ ln 2-2 ln 3 (B) 3- ln 3 (C) 2 ln 3+ ln 2 (D) 1

Tardigrade Admin · Accepted Answer

I =∫ limits-∞0 e x dx +∫ limits0 ln 2( e x -1) dx +∫ limits ln 2 ln 3( e x -2) dx =. e x |-∞ 0+ e x -. x |0 ln 2+ e x -.2 x | ln 2 ln 3 =1+(2- ln 2)-1+(3-2 ln 3)-(2-2 ln 2)=2- ln 2+1-2 ln 3+2 ln 2 =3+ ln 2-2 ln 3

Q. The value of the definite integral $- \infty \int l n 3 {e^{x}} d x$ equals
[Note: ${y}$ denotes fractional part of $y$ .]

$3 + ln 2 - 2 ln 3$

$3 - ln 3$

$2 ln 3 + ln 2$

1

$I = - \infty \int 0 e^{x} d x + 0 \int l n 2 (e^{x} - 1) d x + l n 2 \int l n 3 (e^{x} - 2) d x = e^{x} ∣_{- \infty}^{0} + e^{x} - x ∣_{0}^{l n 2} + e^{x} - 2 x ∣_{l n 2}^{l n 3}$
$= 1 + (2 - ln 2) - 1 + (3 - 2 ln 3) - (2 - 2 ln 2) = 2 - ln 2 + 1 - 2 ln 3 + 2 ln 2$
$= 3 + ln 2 - 2 ln 3$

Q. The value of the definite integral −∞∫ln3​{ex}dx equals [Note: {y} denotes fractional part of y.]

3+ln2−2ln3

3−ln3

2ln3+ln2

1

I=−∞∫0​exdx+0∫ln2​(ex−1)dx+ln2∫ln3​(ex−2)dx=ex∣−∞0​+ex−x∣0ln2​+ex−2x∣ln2ln3​ =1+(2−ln2)−1+(3−2ln3)−(2−2ln2)=2−ln2+1−2ln3+2ln2 =3+ln2−2ln3

Q. The value of the definite integral $- \infty \int l n 3 {e^{x}} d x$ equals
[Note: ${y}$ denotes fractional part of $y$ .]

$3 + ln 2 - 2 ln 3$

$3 - ln 3$

$2 ln 3 + ln 2$

$I = - \infty \int 0 e^{x} d x + 0 \int l n 2 (e^{x} - 1) d x + l n 2 \int l n 3 (e^{x} - 2) d x = e^{x} ∣_{- \infty}^{0} + e^{x} - x ∣_{0}^{l n 2} + e^{x} - 2 x ∣_{l n 2}^{l n 3}$
$= 1 + (2 - ln 2) - 1 + (3 - 2 ln 3) - (2 - 2 ln 2) = 2 - ln 2 + 1 - 2 ln 3 + 2 ln 2$
$= 3 + ln 2 - 2 ln 3$