Question

Use the method of quadrature to estimate the area under the curve y=-x^2+5 and above the x-axis from x=0 to x=2.

Answer 1

I'm a calculus teacher at the high school level and I have never heard of quadrature.  I teach from an AP approved college-level text so...I'll do it using the antiderivative!  Setting that up to integrate we have $- \int\limits^2_0 { (x^2-5)} \, dx$.  If you notice, I pulled the negative out in front of the integral and changed the signs in the function.  So it could really be rewritten as $-1 \int\limits^2_0 {(x^2-5)} \, dx$.  Integrating we have $-1[ \frac{x^3}{3} -5x]$ from 0 to 2.  Subbing in those bounds, we have $-1[( \frac{2^3}{3}-5(2))-( \frac{0^3}{3}-5(0))]$ which simplifies to $-1( \frac{8}{3}-10)$.  Distributing the -1 in gives us $10- \frac{8}{3}= \frac{22}{3}$

Answer 2

Answer:

Area =$\frac{22}{3}$

Step-by-step explanation:

Given is a curve

$y=-x^2+5$

we have to find the area of this curve above x axis from 0 to 2

By method of quadrature we know that area of a curve above x axis is given by,

$\int\limits^a_bf( {x} )\, dx$

Here a= 0 and b =2

Substitute to get

Area = $\int\limits^0_2 {-x^2+5} \, dx \\=\frac{-x^3}{3} +5x$

Substituting limits

Area = $\frac{-8}{3} +10 =\frac{22}{3}$