The area bounded by the curve y = loge x and x-axis and the straight line x = e is ___________ .

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KollegeApply · Accepted Answer

1 sq. units The point of intersection of the curve and the straight line is A(e, 1). Therefore, the area of the required region ABC, A = \int_0^1 \left( x_1 - x_2 ight) d y ..........\left(	ext{ where, }x_1 = e	ext{ and }x_2 = e^y ight) = \int_0^1 \left( e - e^y ight) d y = \left[ ey - e^y ight]_0^1 = \left\{ e\left( 1 ight) - e^\left( 1 ight) ight\} - \left\{ e\left( 0 ight) - e^\left( 0 ight) ight\} = e - e + 1 = 1	ext{ square unit } [Solution diagram: 43927_solution_1.png]

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