The area enclosed by the curves y= log e(x+ .e2), x= log e((2/y)) and x= log e 2, above the line y =1 is

Question

The area enclosed by the curves y= log e(x+ .e2), x= log e((2/y)) and x= log e 2, above the line y =1 is (A) 2+ e - log e 2 (B) 1+ e - log e 2 (C) e - log e 2 (D) 1+ log e 2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/4ddda8d0ee788ab3f491dc5b836ae37c-.png" /> Required area is =∫ limits e - e 20 ln ( x + e 2)-1 dx +∫ limits0 ln 2 2 e - x -1 dx =1+ e - ln 2

Q. The area enclosed by the curves $y = lo g_{e} (x +$ $e^{2}), x = lo g_{e} (\frac{2}{y})$ and $x = lo g_{e} 2$ , above the line $y = 1$ is

$2 + e - lo g_{e} 2$

$1 + e - lo g_{e} 2$

$e - lo g_{e} 2$

$1 + lo g_{e} 2$

Required area is
$= e - e^{2} \int 0 ln (x + e^{2}) - 1 d x + 0 \int l n 2 2 e^{- x} - 1 d x = 1 + e - ln 2$

Q. The area enclosed by the curves y=loge​(x+ e2),x=loge​(y2​) and x=loge​2, above the line y=1 is

2+e−loge​2

1+e−loge​2

e−loge​2

1+loge​2