The area (in sq.

Question

The area (in sq. units) in the first quadrant bounded by the parabola $y=x^2+1$, the tangent to it at the point $(2,5)$, and the coordinate axes is:

Accepted Answer

At $(2,5)$: $\dfrac{dy}{dx}=4$, so tangent: $y-5=4(x-2) \Rightarrow y=4x-3$, meeting the $x$-axis at $\left(\dfrac34,0\right)$.

$$\text{Area} = \int_0^2\left[(x^2+1)-(4x-3)\right]dx - \frac12\times\frac34\times3$$

$$= \int_0^2(x-2)^2\,dx - \frac98 = \left[\frac{(x-2)^3}{3}\right]_0^2 - \frac98 = \frac83-\frac98 = \frac{37}{24}$$

Solutions · 1