If A is the area in the first quadrant enclosed by the curve C:2x^2-y+1=0, the tangent to …

Question

If A is the area in the first quadrant enclosed by the curve $C:2{x}^{2}-y+1=0$, the tangent to $C$ at the point $(1,3)$ and the line $x+y=1$, then the value of $60A$ is \_\_\_\_\_\_\_\_\_.

Answer: 16

Accepted Answer

$y=2{x}^{2}+1$

Tangent at (1, 3) : $y-3=4(x-1)$

$y=4x-1$

$A=\int _{0}^{1}(2{x}^{2}+1)dx-$ area of $(\DeltaQOT)-$ area of $(\DeltaPQR)+$ area of $(\DeltaQRS)$

$A=(\frac{2}{3}+1)-\frac{1}{2}-\frac{9}{8}+\frac{9}{40}=\frac{16}{60}$

$	herefore $ $60A=16$

Solutions · 1