Let g(x) be the inverse of f ( x ) and f ( x ) = 11 x 5 + 7 x 3 + x + 1 , then area bounded by g(x), x-axis and x = 20 , is -

Question

TestHub · Accepted Answer

f is strictly increasing, f(0)=1 and f(1)=20. Reflecting the inverse-curve region gives: $f^{'} (x) = 55 x^{4} + 21 x^{2} + 1 > 0. A = \int_{0}^{1} [20 - f (x)] d x = [19 x - \frac{11 x ^{6}}{6} - \frac{7 x ^{4}}{4} - \frac{x ^{2}}{2}]_{0}^{1} = 19 - \frac{11}{6} - \frac{7}{4} - \frac{1}{2} = \frac{179}{12} .$

Mathematics - Area Under Curves Question with Solution | TestHub

Options:

Doubts & Discussion