Applications of Integration

$\int_0^4 \left( \frac{\sqrt{3}}{4} y^2 \right) , dx$

$\int_0^4 (y)^2 , dx$

$\int_0^4 \left( \frac{y}{3} \right)^3 , dx$

$\int_0^4 \left( \frac{y}{2} \right)^2 , dx$

A two-dimensional object with cross-shaped sections

A three-dimensional object with circular cross sections

A three-dimensional object that can be divided into smaller two-dimensional shapes

A two-dimensional object that can be divided into smaller three-dimensional shapes

$\int_{e^{-\ln(1)}}^{e^{-\ln(4)}} (e^{-x}-e^{-2x})dx$

$\int_{\ln(1)}^{\ln(4)} e^{-5x}dx$

\int_{\ln(1)}^{\ln(4)} (e^{-xB}\nB{-xE9}}

$\int_{\ln(1)}^{\ln(4)} 4(e^{-x}-e^{-2x})dx$

$\pi r$

$\frac{1}{2} \pi r$

$\frac{1}{2} \pi r^2$

$\pi r^2$

Correct answer intentionally omitted until prompted.

Correct answer intentionally omitted until prompted.

Correct answer intentionally omitted until prompted.

$\pi\int_{-\pi/4} ^ {\pi /4} [\sin(x)]^{*}dx$

$\int_0^1 \frac{\sqrt{3}}{4}(x^2)^2 dx$

$\int_0^1 (x^4) dx$

$\int_0^1 \frac{\sqrt{3}}{4}(2x)^2 dx$

$\int_0^1 \pi(x^2)^2 dx$

6

3

12

9

The function that describes the cross section

The area of one cross section

The range of x or y values for which the cross section is defined

The width or height of the slice

$\int_0^1 \pi(x^{3}-x) dx$

$\int_0^1 \frac{\pi}{8}(x^{3}-x)^{2} dx$

$\int_0^{1} \frac{\pi}{4}(x - x^{3})^{2} dx$

$\int_0^1 \frac{\pi}{8}(x+x^{3})^{2} dx$

Approximate by taking the sum of areas of squares at midpoints multiplied by a small $\Delta x$ interval.

Use Riemann sums to approximate each square area and then take their limit as $\Delta x$ approaches zero.

Apply geometry formulas for a cube with side length $\sqrt{x}$ and multiply it by 4.

Using integration, apply the formula $V = \int_{a}^{b} A(x) dx$ where $A(x) = (\sqrt{x})^2$.