Techniques of antidifferentiation - AP Calculus AB

Card 1 of 1166

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

Evaluate the integral $\int \frac{x(5x^3 +4)}{x^5 + 2x^2}dx$

Tap to reveal answer

Answer

This integral can be evaluated using u-substitution. There aren't any groups raised to an exponent so we won't try the power rule for integration. Instead we notice that the integral is still a giant fraction. We will try to match this basic integral form: $\int \frac{du}{u}= \ln|u| + C$

Let's make the denominator our .

Differentiate to get .

This may not look like the numerator, but if we factor out an x, it will.

Now we have all the pieces matched perfectly. Let's make to substitution.

$\int \frac{x(5x^3 +4)}{x^5 + 2x^2}dx$

$\int \frac{du}{u}$

Applying the basic integral form from before, we get

$\ln|u|+C$

Putting everything back in terms of x, we get

$\ln|x^5 + 2x^2| +C$ .

This is the correct answer.

← Didn't Know|Knew It →

Question

Determine the value of $\int_{0}^{2}\sqrt{9x^2+12x+4} dx$ .

Tap to reveal answer

Answer

We can factor the equation inside the square root:

$\\int_0^2\sqrt{9x^2+12x+4}dx \\\int_0^2\sqrt{(3x+2)^2}dx \\\int_0^2 (3x+2)dx$

From here, increase each term's exponent by one and divide the term by the new exponent.

$\frac{3x^2}{2}+2x\big|_0^2$

Now, substitute in the upper bound into the function and subtract the lower bound function value from it.

$\\frac{3(2)^2}{2}+2(2)-\left(\frac{3(0)^2}{2}+2(0)\right) \\\frac{3(4)}{2}+4 \\\frac{12}{2}+4 \\6+4 \\10$

Therefore,

$\int_{0}^{2}\sqrt{(3x+2)^2}dx=\int_{0}^{2}(3x+2)dx=10$

← Didn't Know|Knew It →

Question

Consider the function

Find the minimum of the function on the interval $\left [ 0,1 \right ]$ .

Tap to reveal answer

Answer

To find potential minima of the function, take the first derivative of using the power rule.

Set the derivative to 0:

We solve for to obtain and then plug in 0.5 into the original function to obtain the answer of

We can double check that is indeed a minimum by using the second derivative test

which means the function is concave up, so that the point we found is a minimum.

← Didn't Know|Knew It →

Question

What is the local minimum of when $-1\leq x\leq 1$ ?

Tap to reveal answer

Answer

To find the maximum, we need to look at the first derivative.

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3x^{3-1})+(22x^{2-1})+(05x^{0-1})$

Notice that $(05x^{0-1}) =0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3x^{3-1})+(22x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=(3x^{2})+(22x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x^2+5)=3x^2+4x$

When looking at the first derivative, remember that if the output of this equation is positive, the original function is increasing. If the derivative is negative, then the function is decreasing.

Because we want the MINIMUM, we want to see where the derivative changes from negative to positive.

Notice that has a root when . In fact, it changes from negative to positive at that particular point. This is the local minimum in the interval $-1 \leq x \leq 1$ .

← Didn't Know|Knew It →

Question

Find the x coordinate of the minimum of on the interval $\left [ -2, 2\right ]$ .

Tap to reveal answer

Answer

First, find the derivative, which is:

Set that equal to 0 to find your critical points:

$9x^2-4x=0; x=0, \frac{4}{9}$

Evaluate f at each critical point and endpoint over \[-2, 2\]:

$f(\frac{4}{9})=-\frac{32}{243}$

The least of those numbers is the minimum. Because f(-2)=-32, that is the minimum.

← Didn't Know|Knew It →

Question

Integrate,

$\int[x^2 +\sin(x)-3]dx$

Tap to reveal answer

Answer

Integrate

$\int[x^2 +\sin(x)-3]dx$

1) Apply the sum rule for integration,

$\int x^2 dx +\int \sin(x)dx -\int 3dx$

2) Integrate each individual term and include a constant of integration,

$\frac{1}{3}x^3 - \cos(x) - 3x +C$

Further Discussion

Since indefinite integration is essentially a reverse process of differentiation, check your result by computing its' derivative.

$\frac{d}{dx}\left(\frac{1}{3}x^3-\cos(x)-3x+C \right )=x^2+\sin(x) -3$

This is the same function we integrated, which confirms our result. Also, because the derivative of a constant is always zero, we must include "C" in our result since any constant added to any function will produce the same derivative.

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[cos(ax)]=\frac{sin(ax)}{a}\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=\frac{(27sin(x + 1))}{8}|_{8}^{11.5}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=\frac{(27sin((11.5) + 1))}{8}-(\frac{(27sin((8) + 1))}{8})\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=-1.61\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{27}^{31}(\frac{(8x)}{51})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int x^{a}=\frac{x^{a+1}}{a+1}\&\int_{27}^{31}(\frac{(8x)}{51})dx=\frac{(4x^{2})}{51}|_{27}^{31}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{27}^{31}(\frac{(8x)}{51})dx=\frac{(4(31)^{2})}{51}-(\frac{(4(27)^{2})}{51})\&\int_{27}^{31}(\frac{(8x)}{51})dx=18.20\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx=\frac{(40\cdot 2^{(\frac{x}{4})})}{(7ln(2))}|_{-6}^{-3}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx=\frac{(40\cdot 2^{(\frac{(-3)}{4})})}{(7ln(2))}-(\frac{(40\cdot 2^{(\frac{(-6)}{4})})}{(7ln(2))})\&\int_{-6}^{-3}(\frac{(10\cdot 2^{(\frac{x}{4})})}{7})dx=1.99\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{-24}^{-19}(\frac{59}{(9x)})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int\frac{a}{u}=aln(u)=ln(u^{a})\&\int_{-24}^{-19}(\frac{59}{(9x)})dx=\frac{(59ln(x))}{9}|_{-24}^{-19}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{-24}^{-19}(\frac{59}{(9x)})dx=\frac{(59ln((-19)))}{9}-(\frac{(59ln((-24)))}{9})\&\int_{-24}^{-19}(\frac{59}{(9x)})dx=-1.53\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{-6}^{-2}(\frac{11}{59})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int x^{a}=\frac{x^{a+1}}{a+1}\&\int_{-6}^{-2}(\frac{11}{59})dx=\frac{(11x)}{59}|_{-6}^{-2}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{-6}^{-2}(\frac{11}{59})dx=\frac{(11(-2))}{59}-(\frac{(11(-6))}{59})\&\int_{-6}^{-2}(\frac{11}{59})dx=0.75\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[cos(ax)]=\frac{sin(ax)}{a}\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx=\frac{(9sin(x + 2))}{11}|_{-16}^{-14}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx=\frac{(9sin((-14) + 2))}{11}-(\frac{(9sin((-16) + 2))}{11})\&\int_{-16}^{-14}(\frac{(9cos(x + 2))}{11})dx=1.25\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{-6}^{-2.5}(10e^{(x)})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[e^{ax}]=\frac{e^{ax}}{a}\&\int_{-6}^{-2.5}(10e^{(x)})dx=10e^{(x)}|_{-6}^{-2.5}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{-6}^{-2.5}(10e^{(x)})dx=10e^{((-2.5))}-(10e^{((-6))})\&\int_{-6}^{-2.5}(10e^{(x)})dx=0.80\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{7.5}^{12.5}(\frac{3^{(\frac{x}{3})}}{4})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\&\int_{7.5}^{12.5}(\frac{3^{(\frac{x}{3})}}{4})dx=\frac{(3\cdot 3^{(\frac{x}{3})})}{(4ln(3))}|_{7.5}^{12.5}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{7.5}^{12.5}(\frac{3^{(\frac{x}{3})}}{4})dx=\frac{(3\cdot 3^{(\frac{(12.5)}{3})})}{(4ln(3))}-(\frac{(3\cdot 3^{(\frac{(7.5)}{3})})}{(4ln(3))})\&\int_{7.5}^{12.5}(\frac{3^{(\frac{x}{3})}}{4})dx=55.77\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{14}^{15}(\frac{2^{(\frac{x}{3})}}{20})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\&\int_{14}^{15}(\frac{2^{(\frac{x}{3})}}{20})dx=\frac{(3\cdot 2^{(\frac{x}{3})})}{(20ln(2))}|_{14}^{15}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{14}^{15}(\frac{2^{(\frac{x}{3})}}{20})dx=\frac{(3\cdot 2^{(\frac{(15)}{3})})}{(20ln(2))}-(\frac{(3\cdot 2^{(\frac{(14)}{3})})}{(20ln(2))})\&\int_{14}^{15}(\frac{2^{(\frac{x}{3})}}{20})dx=1.43\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{9}^{12.5}(\frac{(20cos(3x))}{3})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[cos(ax)]=\frac{sin(ax)}{a}\&\int_{9}^{12.5}(\frac{(20cos(3x))}{3})dx=\frac{(20sin(3x))}{9}|_{9}^{12.5}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{9}^{12.5}(\frac{(20cos(3x))}{3})dx=\frac{(20sin(3(12.5)))}{9}-(\frac{(20sin(3(9)))}{9})\&\int_{9}^{12.5}(\frac{(20cos(3x))}{3})dx=-2.56\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{10}^{15}(\frac{51}{(2x)})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int x^{a}=\frac{x^{a+1}}{a+1}\&\int_{10}^{15}(\frac{51}{(2x)})dx=\frac{(51ln(x))}{2}|_{10}^{15}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{10}^{15}(\frac{51}{(2x)})dx=\frac{(51ln((15)))}{2}-(\frac{(51ln((10)))}{2})\&\int_{10}^{15}(\frac{51}{(2x)})dx=10.34\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\text{Evaluate the integral: }\&\int_{-12.5}^{-11}(\frac{23}{(4\cdot 3^{(\frac{x}{2})})})dx\end{align}$

Tap to reveal answer

Answer

$\begin{align}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\&\int_{-12.5}^{-11}(\frac{23}{(4\cdot 3^{(\frac{x}{2})})})dx=-\frac{23}{(2\cdot 3^{(\frac{x}{2})}ln(3))}|_{-12.5}^{-11}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{-12.5}^{-11}(\frac{23}{(4\cdot 3^{(\frac{x}{2})})})dx=-\frac{23}{(2\cdot 3^{(\frac{(-11)}{2})}ln(3))}-(-\frac{23}{(2\cdot 3^{(\frac{(-12.5)}{2})}ln(3))})\&\int_{-12.5}^{-11}(\frac{23}{(4\cdot 3^{(\frac{x}{2})})})dx=5637.19\end{align}$

← Didn't Know|Knew It →

Question

Given , find the general form for the antiderivative .

Tap to reveal answer

Answer

To answer this, we will need to FOIL our function first.

Now can find the antiderivatives of each of these three summands using the power rule.

$f(x) = \frac{4}{3}x^3 +10x^2 +25x+C$ (Don't forget )!

← Didn't Know|Knew It →

Question

Compute the following integral:

$\int 6t^5-12t+6e^t-sin(t)+14cos(t)dt$

Tap to reveal answer

Answer

Compute the following integral:

$\int 6t^5-12t+6e^t-sin(t)+14cos(t)dt$

Now, we need to recall a few rules.

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

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

$\int sin(x)dx=-cos(x)+c$

$\int cos(x)dx=sin(x)+c$

We can use all these rules to change our original function into its anti-derivative.

$\int 6t^5-12t+6e^t-sin(t)+14cos(t)dt$

We can break this up into separate integrals for each term, and apply our rules individually.

$\int 6t^5dt-\int 12tdt+\int 6e^tdt-\int sin(t)dt+\int 14cos(t)dt$

The first two integrals can be found using rule 2

$\int 6t^5dt-\int 12tdt=\frac{6t^6}{6}+c-\frac{12t^2}{2}+c=t^6-6t^2+c$

Next, let's tackle the middle integral:

$\int 6e^tdt=6e^t +c$

Then the "sine" integral

$-\int sin(t)=- (-cos(t))+c=cos(t)+c$

And finally, the cosine integral.

$\int 14cos(t)dt=14sin(t)+c$

Now, we can put all of this together to get:

Note that we only have 1 c, because the c is just a constant.

← Didn't Know|Knew It →

0

Knew It