Question Number 102313 by bemath last updated on 08/Jul/20

Answered by 1549442205 last updated on 08/Jul/20

The abscissa of intersection point of the curve y=(4/(√x))−(4/x) and x−axis being the root of the eqs.(4/(√x))−(4/x)=0 ⇔x−(√x)=0(x≠0)⇔x=1.Hence, S=∫_0 ^1 e^(1−(1/4)x) dx+∫_1 ^4 [e^(1−(1/4)x) −((4/(√x))−(4/x))]dx ∫_0 ^4 e^(1−(1/4)x) dx+∫_1 ^4 (4/x)dx−∫_1 ^4 (4/(√x))dx =−4e^(1−(1/4)x) ∣_0 ^4 +4lnx∣_1 ^4 −8(√(x )) ∣_1 ^4 =−4−(−4e)+4(ln4−0)−8(2−1) =4e+8ln2−12

Commented byI want to learn more last updated on 08/Jul/20

Thanks sir

Answered by bemath last updated on 08/Jul/20

i got Area = ∫_0 ^4 e^(1−(x/4)) dx −∫_1 ^4 (4x^(−1/2) −4x^(−1) )dx = −4∫_0 ^4 e^(1−(x/4)) d(1−(x/4))−{8(√x)−4ln(x)}_1 ^4 =−4{e^(1−(x/4)) }_0 ^4 −{(16−4ln(4)−8} =−4(1−e)+4ln(4)−8 =4e + 4ln(4)−12 ≈ 4.4183

Commented byI want to learn more last updated on 08/Jul/20

Thanks sir