Question

Consider the curve $x=y^4-5y^2+4$. It represents following two explicit functions: $f:\left[-\frac{9}{4},4\right]\rightarrow\left[0,\sqrt{\frac{5}{2}}\right];y=f(x)$ $g:\left[-\frac{9}{4},\infty\right)\rightarrow\left[\sqrt{\frac{5}{2}},\infty\right);y=g(x)$ Let $A_1$ be area bounded by $y=f(x)$ & $xy=0$ as x varies from 0 to 4. $A_2$ be area bounded by $y=g(x), x^2-16x+63=0$ & $y=0$. $A_3$ be the area bounded by tangent to the curve $y=f(x)$ at $P(-2, \sqrt{2})$ and co-ordinate axes. On the basis of above information answer the following:

• A. $\frac{38}{15}$
• B. $\frac{36}{15}$
• C. $\frac{88}{15}$
• D. 3

Answer 1

Answer: (A)

$A_1$ is equal to :-

Answer 2

The area $A_1$ is bounded by $y=f(x)$, $x=0$, $y=0$, and $x=4$. We can calculate this area by integrating $x$ with respect to $y$ from $y=0$ to $y=1$. The equation of the curve is $x = y^4 - 5y^2 + 4$. $A_1 = \int_{0}^{1} (y^4 - 5y^2 + 4) \, dy$ $A_1 = \left[ \frac{y^5}{5} - \frac{5y^3}{3} + 4y \right]_{0}^{1}$ $A_1 = \frac{1}{5} - \frac{5}{3} + 4 = \frac{3 - 25 + 60}{15} = \frac{38}{15}$