Question

Find the area inside function ^(2x^3 + 5 x^2 – 4)⁄x^2 from x = 1 to a.

(a) ^a^2⁄2 + 5a – 4ln(a)

(b) ^a^2⁄2 + 5a – 4ln(a) – ^11⁄2

(c) ^a^2⁄2 + 4ln(a) – ^11⁄2

(d) ^a^2⁄2 + 5a – ^11⁄2

I got this question in a job interview.

This intriguing question comes from Double Integrals in chapter Multiple Integrals of Engineering Mathematics

Answer 1

Correct answer is (b) ^a^2⁄2 + 5a – 4ln(a) – ^11⁄2

To explain: Add constant automatically

Given,

\(f(x) = \frac{(2x^3+5x^2-4)}{x^2}\)

Integrating it we get, F(x) = ^x^2⁄2 + 5x – 4ln(x)

Hence, area under, x = 1 to a, is

F(a) – F(1) = ^a^2⁄2 + 5a – 4ln(a) – ^1⁄2 – 5 = ^a^2⁄2 + 5a – 4ln(a) – ^11⁄2.