Initial Conditions - AP Calculus BC

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The temperature of an oven is increasing at a rate degrees Fahrenheit per miniute for $0\le t\le10$ minutes. The initial temperature of the oven is degrees Fahrenheit.

What is the temperture of the oven at ? Round your answer to the nearest tenth.

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Integrating over an interval will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral

$\int_${0}^{5}$$2e^{0.5t}$dt$

to find the change in temperature that occurs during the first five minutes.

A substitution is useful in this case. Let $u=0.5t, du=0.5dt, \ and\ 2du=dt$ . We should also express the limits of integration in terms of . When , and when Making these substitutions leads to the integral

$\int_${0}^{2.5}$$4e^udu$$ .

To evaluate this, you must know the antiderivative of an exponential function.

In general,

$\int $e^xdx$$=e^x$+c$ .

Therefore,

$\int_${0}^{2.5}$$4e^udu$$=4e^u$|_${0}^{2.5}$$=4e^{2.5}$-4\approx 44.7$ .

This tells us that the temperature rose by approximately degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at minutes is

degrees.

Find the equation for the velocity of a particle if the acceleration of the particle is given by:

and the velocity at time of the particle is .

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In order to find the velocity function, we must integrate the accleration function:

$\int $(16t^2$$+48)dt=\frac{16}{3}$t^3$+48t+C$

We used the rule

$\int $x^ndx$$=\frac{1}{n+1}$x^{n+1}$+C$

to integrate.

Now, we use the initial condition for the velocity function to solve for C. We were told that

so we plug in zero into the velocity function and solve for C:

C is therefore 30.

Finally, we write out the velocity function, with the integer replacing C:

Find the work done by gravity exerting an acceleration of for a block down from its original position with no initial velocity.

Remember that

$W=\int_${a}^{b}$F(x)dx$ , where is a force measured in , is work measured in , and and are initial and final positions respectively.

Tap to see back →

The force of gravity is proportional to the mass of the object and acceleration of the object.

$F=ma= $(10-10)$\frac{kgm}{s^2$}$= -100N$

Since the block fell down 5 meters, its final position is and initial position is .

$\ W=\int_${0}^{-5}$-100dx\ \=\frac{-100x}{2}$|_${0}^{-5}$ \ \=-50x|_${0}^{-5}$\ \=-50(-5)--50(0)\ \=250J$

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

Tap to see back →

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

To solve this problem,we need to evaluate the given integral, then solve for our constant of integration.

Let's begin by recalling the following integration rules:

$1) \int $x^ndx$$=\frac{x^{n+1}$$}{n+1}+c$

$2) \int $e^xdx$$=e^x$+c$

Using these two, we can integrate f(x)

SO, we get:

We are almost there, but we need to find c. To do so, plug in our initial conditions and solve:

So, our answer is:

The temperature of an oven is increasing at a rate degrees Fahrenheit per miniute for $0\le t\le10$ minutes. The initial temperature of the oven is degrees Fahrenheit.

What is the temperture of the oven at ? Round your answer to the nearest tenth.

Tap to see back →

Integrating over an interval will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral

$\int_${0}^{5}$$2e^{0.5t}$dt$

to find the change in temperature that occurs during the first five minutes.

A substitution is useful in this case. Let $u=0.5t, du=0.5dt, \ and\ 2du=dt$ . We should also express the limits of integration in terms of . When , and when Making these substitutions leads to the integral

$\int_${0}^{2.5}$$4e^udu$$ .

To evaluate this, you must know the antiderivative of an exponential function.

In general,

$\int $e^xdx$$=e^x$+c$ .

Therefore,

$\int_${0}^{2.5}$$4e^udu$$=4e^u$|_${0}^{2.5}$$=4e^{2.5}$-4\approx 44.7$ .

This tells us that the temperature rose by approximately degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at minutes is

degrees.

Find the equation for the velocity of a particle if the acceleration of the particle is given by:

and the velocity at time of the particle is .

Tap to see back →

In order to find the velocity function, we must integrate the accleration function:

$\int $(16t^2$$+48)dt=\frac{16}{3}$t^3$+48t+C$

We used the rule

$\int $x^ndx$$=\frac{1}{n+1}$x^{n+1}$+C$

to integrate.

Now, we use the initial condition for the velocity function to solve for C. We were told that

so we plug in zero into the velocity function and solve for C:

C is therefore 30.

Finally, we write out the velocity function, with the integer replacing C:

Find the work done by gravity exerting an acceleration of for a block down from its original position with no initial velocity.

Remember that

$W=\int_${a}^{b}$F(x)dx$ , where is a force measured in , is work measured in , and and are initial and final positions respectively.

Tap to see back →

The force of gravity is proportional to the mass of the object and acceleration of the object.

$F=ma= $(10-10)$\frac{kgm}{s^2$}$= -100N$

Since the block fell down 5 meters, its final position is and initial position is .

$\ W=\int_${0}^{-5}$-100dx\ \=\frac{-100x}{2}$|_${0}^{-5}$ \ \=-50x|_${0}^{-5}$\ \=-50(-5)--50(0)\ \=250J$

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

Tap to see back →

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

To solve this problem,we need to evaluate the given integral, then solve for our constant of integration.

Let's begin by recalling the following integration rules:

$1) \int $x^ndx$$=\frac{x^{n+1}$$}{n+1}+c$

$2) \int $e^xdx$$=e^x$+c$

Using these two, we can integrate f(x)

SO, we get:

We are almost there, but we need to find c. To do so, plug in our initial conditions and solve:

So, our answer is:

The temperature of an oven is increasing at a rate degrees Fahrenheit per miniute for $0\le t\le10$ minutes. The initial temperature of the oven is degrees Fahrenheit.

What is the temperture of the oven at ? Round your answer to the nearest tenth.

Tap to see back →

Integrating over an interval will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral

$\int_${0}^{5}$$2e^{0.5t}$dt$

to find the change in temperature that occurs during the first five minutes.

A substitution is useful in this case. Let $u=0.5t, du=0.5dt, \ and\ 2du=dt$ . We should also express the limits of integration in terms of . When , and when Making these substitutions leads to the integral

$\int_${0}^{2.5}$$4e^udu$$ .

To evaluate this, you must know the antiderivative of an exponential function.

In general,

$\int $e^xdx$$=e^x$+c$ .

Therefore,

$\int_${0}^{2.5}$$4e^udu$$=4e^u$|_${0}^{2.5}$$=4e^{2.5}$-4\approx 44.7$ .

This tells us that the temperature rose by approximately degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at minutes is

degrees.

Find the equation for the velocity of a particle if the acceleration of the particle is given by:

and the velocity at time of the particle is .

Tap to see back →

In order to find the velocity function, we must integrate the accleration function:

$\int $(16t^2$$+48)dt=\frac{16}{3}$t^3$+48t+C$

We used the rule

$\int $x^ndx$$=\frac{1}{n+1}$x^{n+1}$+C$

to integrate.

Now, we use the initial condition for the velocity function to solve for C. We were told that

so we plug in zero into the velocity function and solve for C:

C is therefore 30.

Finally, we write out the velocity function, with the integer replacing C:

Find the work done by gravity exerting an acceleration of for a block down from its original position with no initial velocity.

Remember that

$W=\int_${a}^{b}$F(x)dx$ , where is a force measured in , is work measured in , and and are initial and final positions respectively.

Tap to see back →

The force of gravity is proportional to the mass of the object and acceleration of the object.

$F=ma= $(10-10)$\frac{kgm}{s^2$}$= -100N$

Since the block fell down 5 meters, its final position is and initial position is .

$\ W=\int_${0}^{-5}$-100dx\ \=\frac{-100x}{2}$|_${0}^{-5}$ \ \=-50x|_${0}^{-5}$\ \=-50(-5)--50(0)\ \=250J$

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

Tap to see back →

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

To solve this problem,we need to evaluate the given integral, then solve for our constant of integration.

Let's begin by recalling the following integration rules:

$1) \int $x^ndx$$=\frac{x^{n+1}$$}{n+1}+c$

$2) \int $e^xdx$$=e^x$+c$

Using these two, we can integrate f(x)

SO, we get:

We are almost there, but we need to find c. To do so, plug in our initial conditions and solve:

So, our answer is:

The temperature of an oven is increasing at a rate degrees Fahrenheit per miniute for $0\le t\le10$ minutes. The initial temperature of the oven is degrees Fahrenheit.

What is the temperture of the oven at ? Round your answer to the nearest tenth.

Tap to see back →

Integrating over an interval will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral

$\int_${0}^{5}$$2e^{0.5t}$dt$

to find the change in temperature that occurs during the first five minutes.

A substitution is useful in this case. Let $u=0.5t, du=0.5dt, \ and\ 2du=dt$ . We should also express the limits of integration in terms of . When , and when Making these substitutions leads to the integral

$\int_${0}^{2.5}$$4e^udu$$ .

To evaluate this, you must know the antiderivative of an exponential function.

In general,

$\int $e^xdx$$=e^x$+c$ .

Therefore,

$\int_${0}^{2.5}$$4e^udu$$=4e^u$|_${0}^{2.5}$$=4e^{2.5}$-4\approx 44.7$ .

This tells us that the temperature rose by approximately degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at minutes is

degrees.

Find the equation for the velocity of a particle if the acceleration of the particle is given by:

and the velocity at time of the particle is .

Tap to see back →

In order to find the velocity function, we must integrate the accleration function:

$\int $(16t^2$$+48)dt=\frac{16}{3}$t^3$+48t+C$

We used the rule

$\int $x^ndx$$=\frac{1}{n+1}$x^{n+1}$+C$

to integrate.

Now, we use the initial condition for the velocity function to solve for C. We were told that

so we plug in zero into the velocity function and solve for C:

C is therefore 30.

Finally, we write out the velocity function, with the integer replacing C:

Find the work done by gravity exerting an acceleration of for a block down from its original position with no initial velocity.

Remember that

$W=\int_${a}^{b}$F(x)dx$ , where is a force measured in , is work measured in , and and are initial and final positions respectively.

Tap to see back →

The force of gravity is proportional to the mass of the object and acceleration of the object.

$F=ma= $(10-10)$\frac{kgm}{s^2$}$= -100N$

Since the block fell down 5 meters, its final position is and initial position is .

$\ W=\int_${0}^{-5}$-100dx\ \=\frac{-100x}{2}$|_${0}^{-5}$ \ \=-50x|_${0}^{-5}$\ \=-50(-5)--50(0)\ \=250J$

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

Tap to see back →

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

To solve this problem,we need to evaluate the given integral, then solve for our constant of integration.

Let's begin by recalling the following integration rules:

$1) \int $x^ndx$$=\frac{x^{n+1}$$}{n+1}+c$

$2) \int $e^xdx$$=e^x$+c$

Using these two, we can integrate f(x)

SO, we get:

We are almost there, but we need to find c. To do so, plug in our initial conditions and solve:

So, our answer is:

The temperature of an oven is increasing at a rate degrees Fahrenheit per miniute for $0\le t\le10$ minutes. The initial temperature of the oven is degrees Fahrenheit.

What is the temperture of the oven at ? Round your answer to the nearest tenth.

Tap to see back →

Integrating over an interval will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral

$\int_${0}^{5}$$2e^{0.5t}$dt$

to find the change in temperature that occurs during the first five minutes.

A substitution is useful in this case. Let $u=0.5t, du=0.5dt, \ and\ 2du=dt$ . We should also express the limits of integration in terms of . When , and when Making these substitutions leads to the integral

$\int_${0}^{2.5}$$4e^udu$$ .

To evaluate this, you must know the antiderivative of an exponential function.

In general,

$\int $e^xdx$$=e^x$+c$ .

Therefore,

$\int_${0}^{2.5}$$4e^udu$$=4e^u$|_${0}^{2.5}$$=4e^{2.5}$-4\approx 44.7$ .

This tells us that the temperature rose by approximately degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at minutes is

degrees.

Find the equation for the velocity of a particle if the acceleration of the particle is given by:

and the velocity at time of the particle is .

Tap to see back →

In order to find the velocity function, we must integrate the accleration function:

$\int $(16t^2$$+48)dt=\frac{16}{3}$t^3$+48t+C$

We used the rule

$\int $x^ndx$$=\frac{1}{n+1}$x^{n+1}$+C$

to integrate.

Now, we use the initial condition for the velocity function to solve for C. We were told that

so we plug in zero into the velocity function and solve for C:

C is therefore 30.

Finally, we write out the velocity function, with the integer replacing C:

Find the work done by gravity exerting an acceleration of for a block down from its original position with no initial velocity.

Remember that

$W=\int_${a}^{b}$F(x)dx$ , where is a force measured in , is work measured in , and and are initial and final positions respectively.

Tap to see back →

The force of gravity is proportional to the mass of the object and acceleration of the object.

$F=ma= $(10-10)$\frac{kgm}{s^2$}$= -100N$

Since the block fell down 5 meters, its final position is and initial position is .

$\ W=\int_${0}^{-5}$-100dx\ \=\frac{-100x}{2}$|_${0}^{-5}$ \ \=-50x|_${0}^{-5}$\ \=-50(-5)--50(0)\ \=250J$

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

Tap to see back →

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

To solve this problem,we need to evaluate the given integral, then solve for our constant of integration.

Let's begin by recalling the following integration rules:

$1) \int $x^ndx$$=\frac{x^{n+1}$$}{n+1}+c$

$2) \int $e^xdx$$=e^x$+c$

Using these two, we can integrate f(x)

SO, we get:

We are almost there, but we need to find c. To do so, plug in our initial conditions and solve:

So, our answer is: