a million) vital of x^2 y dydx = x^2 * y^2/2 for 2 < y < 4: x^2 * (sixteen/2 - 4/2) = 6x^2 vital of 6x^2 dx = 6x^3/3 = 2x^3 for 0 < x < a million: 2(a million)^3 - 0 = 2 no longer 3. 2) y = x/2 -and- y = sqrt(x) x^2/4 = x, x = 0 & 4 Integrating (sqrt(x) - x/2) dx for x = 2 to 4: 2/3 x^(3/2) - x^2/4. Plug in 4 & 2 and spot what you get. 3) ultimate. person-friendly to make certain with out calculus - that is one sector of a unit circle, so it may be pi/4.
Comments
If there's an extremum in the interior of the region, it will occur at a critical point.
∇f = 2x i + (4y - 2) j = <0, 0> ==> x = 0 and y = 1/2.
The point (0, 1/2) is inside the circle.
On the boundary,
f(x, y) = x² + 2y² - 2y = (x² + y²) + y² - 2y = y² - 2y + 5 = (y - 1)² + 4,
and -√5 ≤ y ≤ √5. The minimum value of this occurs at the vertex where y = 1 and f = 4. The maximum occurs at y = -√5 where f = 10 + 2√5
f(0, 1/2) = -1/2 <----absolute minimum
f(±2, 1) = 4
f(0, -√5) = 10 + 2√(5) <----absolute maximum
a million) vital of x^2 y dydx = x^2 * y^2/2 for 2 < y < 4: x^2 * (sixteen/2 - 4/2) = 6x^2 vital of 6x^2 dx = 6x^3/3 = 2x^3 for 0 < x < a million: 2(a million)^3 - 0 = 2 no longer 3. 2) y = x/2 -and- y = sqrt(x) x^2/4 = x, x = 0 & 4 Integrating (sqrt(x) - x/2) dx for x = 2 to 4: 2/3 x^(3/2) - x^2/4. Plug in 4 & 2 and spot what you get. 3) ultimate. person-friendly to make certain with out calculus - that is one sector of a unit circle, so it may be pi/4.