Math / Calculus

QuestionIf $t$ is in years, and $t=0$ is January 1,2020 , worldwide energy consumption, $r$ , in exajoules ( $10^{18}$ joules) per year, ${ }^{1}$ is modeled by $r=583.9 e^{0.013 t}$ (a) Write a definite integral for the total energy use between the start of 2020 and the start of 2029.
Total energy used $=$ $\square$ $d t$ exajoules

Studdy Solution

STEP 1

What is this asking? We're given a formula for yearly energy consumption and need to set up an integral to find the total energy consumed over a specific time period. Watch out! The problem gives the formula with $t=0$ representing the start of 2020.
Make sure to use the correct values for $t$ when setting up the integral!

STEP 2

1. Determine the time range.
2. Set up the definite integral.

STEP 3

We're looking at the energy consumption between the start of 2020 and the start of 2029.
Since $t=0$ corresponds to the start of 2020, our starting time is $t = \mathbf{0}$ .

STEP 4

Since $t$ is measured in years, the start of 2029 corresponds to $t = \mathbf{9}$ (because 2029 is 9 years after 2020).

STEP 5

The formula $r = 583.9 e^{0.013t}$ gives the rate of energy consumption at any time $t$ .
To find the total energy consumed between $t=0$ and $t=9$ , we need to integrate the rate function over this time interval.

STEP 6

The definite integral representing the total energy consumption is: $\int_{\mathbf{0}}^{\mathbf{9}} 583.9 e^{0.013t} \, dt$ This integral calculates the area under the rate curve between $t=0$ and $t=9$ , which represents the total energy consumed during that period.
The $dt$ represents an infinitesimally small change in time, and we're summing up the energy consumption over all these tiny time intervals.

STEP 7

Total energy used = $\int_{0}^{9} 583.9 e^{0.013t} \, dt$ exajoules

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Studdy Solution

STEP 1

STEP 2

1. Determine the time range.
2. Set up the definite integral.

STEP 3

We're looking at the energy consumption between the start of 2020 and the start of 2029.
Since $t=0$ corresponds to the start of 2020, our starting time is $t = \mathbf{0}$ .

STEP 4

Since $t$ is measured in years, the start of 2029 corresponds to $t = \mathbf{9}$ (because 2029 is 9 years after 2020).

STEP 5

The formula $r = 583.9 e^{0.013t}$ gives the rate of energy consumption at any time $t$ .
To find the total energy consumed between $t=0$ and $t=9$ , we need to integrate the rate function over this time interval.

STEP 6

STEP 7

Total energy used = $\int_{0}^{9} 583.9 e^{0.013t} \, dt$ exajoules

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Studdy Solution

STEP 1

STEP 2

1. Determine the time range.2. Set up the definite integral.

STEP 3

We're looking at the energy consumption between the start of **2020** and the start of **2029**.Since t=0t=0t=0 corresponds to the start of 2020, our starting time is t=0t = \mathbf{0}t=0.

STEP 4

Since ttt is measured in years, the start of 2029 corresponds to t=9t = \mathbf{9}t=9 (because 2029 is **9** years after 2020).

STEP 5

The formula r=583.9e0.013tr = 583.9 e^{0.013t}r=583.9e0.013t gives the *rate* of energy consumption at any time ttt.To find the *total* energy consumed between t=0t=0t=0 and t=9t=9t=9, we need to integrate the rate function over this time interval.

STEP 6

STEP 7

Total energy used = ∫09583.9e0.013t dt\int_{0}^{9} 583.9 e^{0.013t} \, dt∫09​583.9e0.013tdt exajoules

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1. Determine the time range.
2. Set up the definite integral.

We're looking at the energy consumption between the start of 2020 and the start of 2029.
Since $t=0$ corresponds to the start of 2020, our starting time is $t = \mathbf{0}$ .

Since $t$ is measured in years, the start of 2029 corresponds to $t = \mathbf{9}$ (because 2029 is 9 years after 2020).

The formula $r = 583.9 e^{0.013t}$ gives the rate of energy consumption at any time $t$ .
To find the total energy consumed between $t=0$ and $t=9$ , we need to integrate the rate function over this time interval.

Total energy used = $\int_{0}^{9} 583.9 e^{0.013t} \, dt$ exajoules