Applications of Integration

$\pi \int_{0}^{3} (y^4) dy$

$\pi \int_{0}^{9} \sqrt{x} dx$

$\pi \int_{0}^{9} (x) dx$

$\pi \int_{0}^{3} (9-y^4) dy$

The volume would be doubled

The volume would be multiplied by 4

The volume would be multiplied by 8

The volume would remain unchanged

Differentiation and integration

Derivation

Integration

Differentiation

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

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

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

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

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

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

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

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

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

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

π

1/2π

2π

4π

Function $f(y)$ needs only to be defined at endpoints $a$ and $b$ for integrability purposes.

Function $f(y)$ can have discontinuities but must not have vertical asymptotes on $[a,b]$.

The integral of $f(y)^2$ from $a$ to $b$ exists irrespective of continuity or discontinuity.

Function $f(y)$ must be continuous on $[a,b]$.

Shell method

Disk method

Cross-sectional area method

Washer method