AP Calculus AB - Integrals

$\begin{align*}&\int_{13}^{4}f(x)dx=42\&\int_{17}^{13}f(x)dx=-56\&\int_{17}^{28}f(x)dx=35\&\int_{28}^{33}f(x)dx=-99\&\int_{4}^{48}f(x)dx=-10\&\text{Determine }\int_{48}^{33}f(x)dx\end{align*}$

Explanation

$\begin{align*}&\text{A principle of integrals is one of summation. If the values}\&\text{of an integral across given sets of points is known, they could}\&\text{ be used to find integral values across other points. Imagine}\&\text{a set of points that are in ascending numerical value: a, b,}\&\text{c, and d. It follows that:}\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\&\text{Also, if an integral value is known, the same integral backwards}\&\text{will give a negative of the original value:}\&\int_a^b=-\int_b^a\&\text{Knowing all this, we can calculate }\int_{48}^{33}f(x)dx\&\int_{4}^{48}f(x)dx=-\int_{13}^{4}f(x)dx-\int_{17}^{13}f(x)dx+\int_{17}^{28}f(x)dx+\int_{28}^{33}f(x)dx-\int_{48}^{33}f(x)dx\&-\int_{48}^{33}f(x)dx=\int_{4}^{48}f(x)dx+\int_{13}^{4}f(x)dx+\int_{17}^{13}f(x)dx-\int_{17}^{28}f(x)dx-\int_{28}^{33}f(x)dx\&-\int_{48}^{33}f(x)dx=-10+(42)+(-56)-(35)-(-99)\&\int_{48}^{33}f(x)dx=-40\&\end{align*}$

$\begin{align*}&\int_{10}^{19}f(x)dx=-40\&\int_{19}^{30}f(x)dx=1\&\int_{42}^{30}f(x)dx=-12\&\int_{4}^{42}f(x)dx=-19\&\text{Use the above values to find }\int_{10}^{4}f(x)dx\end{align*}$

Explanation

$\begin{align*}&\text{A principle of integrals is one of summation. If the values}\&\text{of an integral across given sets of points is known, they could}\&\text{ be used to find integral values across other points. Imagine}\&\text{a set of points that are in ascending numerical value: a, b,}\&\text{c, and d. It follows that:}\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\&\text{Also, if an integral value is known, the same integral backwards}\&\text{will give a negative of the original value:}\&\int_a^b=-\int_b^a\&\text{Knowing all this, we can calculate }\int_{10}^{4}f(x)dx\&\int_{4}^{42}f(x)dx=-\int_{10}^{4}f(x)dx+\int_{10}^{19}f(x)dx+\int_{19}^{30}f(x)dx-\int_{42}^{30}f(x)dx\&-\int_{10}^{4}f(x)dx=\int_{4}^{42}f(x)dx-\int_{10}^{19}f(x)dx-\int_{19}^{30}f(x)dx+\int_{42}^{30}f(x)dx\&-\int_{10}^{4}f(x)dx=-19-(-40)-(1)+(-12)\&\int_{10}^{4}f(x)dx=-8\&\end{align*}$

A particle at the origin has an initial velocity of . If its acceleration is given by , find the position of the particle after 1 second.

Explanation

In this problem, letting denote the position of the particle and denote the velocity, we know that . Integrating and working backwards we have,

$v(t) = \int a(t)dt = \int (-24t^2 + 6t - 8)dt = -8t^3 + 3t^2 -8t + C$

Plugging in our initial condition, , we see immediately that .

Repeating the process again for , we find that

$s(t) = \int v(t)dt = \int (-8t^3 + 3t^2 -8t +32)dt = -2t^4 + t^3 - 4t^2 + 32t + C$

Plugging in our initial condition, (we started at the origin) we see that . This gives us a final equation

. The problem asks for which is simply

Solve the separable differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{4y}{x^2}, x>0$

and at

$y=2e^{4-\frac{4}{x}}$

$y=2e^{-\frac{4}{x}}$

$y=2e^{-\frac{4}{x}}+C$

$y=Ce^{-\frac{4}{x}}$

Explanation

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

$\frac{dy}{y}=\frac{dx}{x^2}$

Next, we integrate both sides:

$\ln \left | y\right | = -\frac{1}{x}+C$

The integrals were solved using the following rules:

$\int \frac{dx}{x}= \ln \left | x\right |+C$ , $\int x^n dx = \frac{x^{n+1}}{n+1}+C$

The two constants of integration were combined to make a single one.

Now, we exponentiate both sides to solve for y:

$y=e^{-\frac{1}{x}+C}$

Using the properties of exponents, we can rearrange the integration constant:

$y=e^{-\frac{1}{x}+C} = e^{-\frac{1}{x}}\cdot e^C = Ce^{-\frac{1}{x}}$

(The exponential of the constant is itself a constant.)

Using the given condition, we can solve for C:

$2=Ce^{-4}$

Our final answer is

$y=2e^{4-\frac{4}{x}}$

Given that and , solve for . What is the value of ?

Explanation

This is a separable differential equation. The simplest way to approach this is to turn into $\frac{\mathrm{d} y}{\mathrm{d} x}$ , and then by abusing the notation, "multiplying by dx" on both sides.

We then group all the y terms with dy, and all the x terms with dx.

$\frac{1}{2y+2}dy = (x - 2)dx$

Integrating both sides, we find

$\int \frac{1}{2y+2}dy = \frac{\ln|2y+2|}{2} =\int (x - 2)dx = \frac{x^2}{2} - 2x + C$

Here, the first integral is found by using substitution of variables, setting . In addition, we have chosen to only put a +C on the second integral, as if we put it on both, we would just combine them in any case.

To solve for y, we multiply both sides by two and raise e to both sides to get rid of the natural logarithm.

$|2y+2| = e^{x^2 - 4x + C}$

(Note, C was multiplied by two, but it's still just an arbitrary constant. If you prefer, you may call the new C value .)

Now we drop our absolute value signs, and note that we can take out a factor of and stick in front of the right hand side.

$2y+2 =\pm e^Ce^{x^2 - 4x}$

As $\pm e^C$ is just another arbitrary constant, we can relabel this as C, or if you prefer. Solving for y gets us

$y = \frac{Ce^{x^2 - 4x} - 2}{2}$

Next, we plug in our initial condition to solve for C.

$1 =\frac{C - 2}{2}$ ;

Leaving us with a final equation of

$y = \frac{4e^{x^2 - 4x} - 2}{2}$

Plugging in x = 4, we have a final answer,

$y = \frac{4e^{16 - 16} - 2}{2} = \frac{4 - 2}{2} = 1$

$\begin{align*}&\int_{-9}^{0}f(x)dx=62\&\int_{13}^{0}f(x)dx=-46\&\int_{18}^{13}f(x)dx=20\&\text{Use the above values to find }\int_{-9}^{18}f(x)dx\end{align*}$

Explanation

$\begin{align*}&\text{One of the principles of integrals is one of summation. If the}\&\text{values of an integral across certain sets of points is known,}\&\text{this could potentially be used to find integral values at other}\&\text{points. Imagine a set of points that are in ascending numerical}\&\text{value: a, b, c, and d. It follows that:}\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\&\text{Another principle of integrals is that if an integral value}\&\text{is known, to take the same integral backwards will give a negative}\&\text{of the original value:}\&\int_a^b=-\int_b^a\&\text{Knowing this, we can calculate }\int_{-9}^{18}f(x)dx\&\int_{-9}^{18}f(x)dx=\int_{-9}^{0}f(x)dx-\int_{13}^{0}f(x)dx-\int_{18}^{13}f(x)dx\&\int_{-9}^{18}f(x)dx=(62)-(-46)-(20)\&\int_{-9}^{18}f(x)dx=88\&\end{align*}$

Find (dy/dx).

sin(xy) = x + cos(y)

dy/dx = (1 – ycos(xy))/(xcos(xy) + sin(y))

dy/dx = (1 – cos(xy))/(cos(xy) + sin(y))

dy/dx = (xcos(xy) + sin(y))/(1 – ycos(xy))

dy/dx = (cos(xy) + sin(y))/(1 – cos(xy))

None of the above

Explanation

The first step of the problem is to differentiate with respect to (dy/dx):

cos(xy)\[(x)(dy/dx) + y(1)\] = 1 – sin(y)(dy/dx)

*Note: When differentiating cos(xy) remember to use the product rule. (xy' + x'y)

Step 2: Clean the differentiated problem up

cos(xy)(x)(dy/dx) + cos(xy)y = 1 – sin(y)(dy/dx)

cos(xy)(x)(dy/dx) + sin(y)(dy/dx) = 1 – cos(xy)y

Step 3: Solve for (dy/dx)

dy/dx = (1 – ycos(xy))/(xcos(xy) + sin(y))

Find the derivative:

Explanation

To find the derivative, multiply the exponent by the coefficent in front of the x term and then decrease the exponent by 1:

$\begin{align*}&\int_{5}^{12}f(x)dx=99\&\int_{26}^{12}f(x)dx=-39\&\int_{0}^{26}f(x)dx=108\&\text{Calculate from the above values }\int_{5}^{0}f(x)dx\end{align*}$

Explanation

$\begin{align*}&\text{A principle of integrals is one of summation. If the values}\&\text{of an integral across given sets of points is known, they could}\&\text{ be used to find integral values across other points. Imagine}\&\text{a set of points that are in ascending numerical value: a, b,}\&\text{c, and d. It follows that:}\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\&\text{Also, if an integral value is known, the same integral backwards}\&\text{will give a negative of the original value:}\&\int_a^b=-\int_b^a\&\text{Knowing all this, we can calculate }\int_{5}^{0}f(x)dx\&\int_{0}^{26}f(x)dx=-\int_{5}^{0}f(x)dx+\int_{5}^{12}f(x)dx-\int_{26}^{12}f(x)dx\&-\int_{5}^{0}f(x)dx=\int_{0}^{26}f(x)dx-\int_{5}^{12}f(x)dx+\int_{26}^{12}f(x)dx\&-\int_{5}^{0}f(x)dx=108-(99)+(-39)\&\int_{5}^{0}f(x)dx=30\&\end{align*}$