Applications of Integration

Differentiation and integration

Derivation

Integration

Differentiation

V = $\int_a^b \frac{\sqrt{3}}{4}s^2 , dx$

V = $\int_a^b \frac{1}{2}\pi r^2 , dx$

V = $\int_a^b s^2 , dx$

V = $\int_a^b \frac{1}{2}s^2 , dx$

Shell method

Disk method

Cross-sectional area method

Washer method

$\int_{f(1)}^{f(5)} (2(y - f^{-1}(y)))^2 dy$

$\int_{f(1)}^{f(5)} (y - f^{-1}(y))^2 dy$

$\int_{f(1)}^{f(5)} 4(y - f^{-1}(y)) dy$

$\int_{f(1)}^{f(5)} (y + f^{-1}(y))^3 dy$

Sum areas of all possible right triangles within bounds y=x and y=$x^2$, then multiply by depth.

Multiply half-base times height for one triangle at point $(a,a^2)$ and scale up linearly along x-axis until point $(b,b^2)$.

Integrate using $V = \int_{a}^{b} A(x) dx$ with $A(x)$ representing the area of an equilateral triangle as a function of $x$.

Calculate volume using cylinder approximation methods between curves y=x and y=$x^2$ substituting triangle bases.

Continuity in $V'(x)$ isn't necessary; only non-negativity matters since areas cannot shrink in size physically.

Function $V'(x)$ may possess finite number of removable discontinuities without affecting overall solid's calculated volume.

The rate-of-change $V'(x)$ can assume discrete values corresponding to each distinct cross-section considered separately.

The rate-of-change described by $V'(x)$ should also be continuous if $V(x)$ is to properly represent volume changes without sudden jumps.

$\int_{0}^{4}(x^2-4)^2dx$

$\int_{0}^{2}(4-x^2)^2dx$

$\int_{0}^{2}\sqrt{4-x^2}dx$

$\int_{0}^{2}(x^4-16)dx$

The volume would be doubled

The volume would be multiplied by 4

The volume would be multiplied by 8

The volume would remain unchanged

Multiplying the area of one cross section by the width or height of the slice

Adding up the volumes of all the slices

Integrating the function for the cross section with respect to x or y

Dividing the area of one cross section by the width or height of the slice

$\int_0^1 (x+x^3) dx$

$\int_0^\frac{1}{\sqrt{x}}(x - x^{5/6})dx$

$\int_0^1 (x-x^3)^2 dx$

$\int_0^\frac{1}{\sqrt[3]{x}} x^{6}dx$