Question

Since YouTube first became available to the public in mid-2005, the rate at which video has been uploaded to the site can be approximated by v(t) = 525,600(0.42t2 + 2.7t − 1) hours of video per year (0.5 ≤ t ≤ 7), where t is time in years since the start of 2005.† Use a definite integral to estimate the total number of hours of video uploaded from the start of 2006 to the start of 2011. (Round your answer to the nearest million hours of video.)

Answer 1

`\displaystyle \int\limits^(6)_(1) {525600(0.42t^2 + 2.7t - 1)} \, dt \approx 38000000`

General Formulas and Concepts:

Calculus

Integration

• Integrals

Integration Rule [Reverse Power Rule]:
`\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C`

Integration Rule [Fundamental Theorem of Calculus 1]:
`\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)`

Integration Property [Multiplied Constant]:
`\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx`

Integration Property [Addition/Subtraction]:
`\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx`

Explanation:

Step 1: Define

Identify

`\displaystyle v(t) = 525600(0.42t^2 + 2.7t - 1) ,\ 0.5 \leq t \leq 7`

Step 2: Find

1. Set up:
   `\displaystyle \int\limits^(6)_(1) {525600(0.42t^2 + 2.7t - 1)} \, dt`
2. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int\limits^(6)_(1) {525600(0.42t^2 + 2.7t - 1)} \, dt = 525600\int\limits^(6)_(1) {(0.42t^2 + 2.7t - 1)} \, dt`
3. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
   `\displaystyle \int\limits^(6)_(1) {525600(0.42t^2 + 2.7t - 1)} \, dt = 525600 \bigg( \int\limits^(6)_(1) {0.42t^2} \, dt + \int\limits^(6)_(1) {2.7t} \, dt - \int\limits^(6)_(1) {1} \, dt \bigg)`
4. [Integrals] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int\limits^(6)_(1) {525600(0.42t^2 + 2.7t - 1)} \, dt = 525600 \bigg( 0.42\int\limits^(6)_(1) {t^2} \, dt + 2.7\int\limits^(6)_(1) {t} \, dt - \int\limits^(6)_(1) {} \, dt \bigg)`
5. [Integrals] Integrate [Integration Rule - Reverse Power Rule]:
   `\displaystyle \int\limits^(6)_(1) {525600(0.42t^2 + 2.7t - 1)} \, dt = 525600 \bigg[ 0.42 \bigg( (t^3)/(3) \bigg) \bigg| \limits^(6)_(1) + 2.7 \bigg( (t^2)/(2) \bigg) \bigg| \limits^(6)_(1) - t \bigg| \limits^(6)_(1) \bigg]`
6. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   `\displaystyle \int\limits^(6.5)_(1.5) {525600(0.42t^2 + 2.7t - 1)} \, dt = 525600 \bigg[ 0.42 \bigg( (215)/(3) \bigg) + 2.7 \bigg( (35)/(2) \bigg) - 5 \bigg]`
7. Simplify:
   `\displaystyle \int\limits^(6)_(1) {525600(0.42t^2 + 2.7t - 1)} \, dt = 38027160`
8. Round:
   `\displaystyle \int\limits^(6)_(1) {525600(0.42t^2 + 2.7t - 1)} \, dt \approx 38000000`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration