Antiderivatives following directly from derivatives of basic functions - AP Calculus AB

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Question

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

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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.

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

Therefore,

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

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Question

Consider the function

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

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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.

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Question

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

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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 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$$+5)=(3x^{2}$$)+(2*2x^{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 $-1 \leq x \leq 1$ .

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Question

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

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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.

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Question

Integrate,

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Answer

Integrate

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,

Further Discussion

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

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.

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Question

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

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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}$

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Question

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

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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}$

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Question

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

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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}$)})}{7})dx=1.99\end{align}$

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Question

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

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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)}$)dx=-1.53\end{align}$

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Question

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

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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}$

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Question

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

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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}$

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Question

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

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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)}$)dx=0.80\end{align}$

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Question

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

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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}$)}}{4})dx=55.77\end{align}$

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Question

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

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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}$)}}{20})dx=1.43\end{align}$

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Question

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

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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}$

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Question

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

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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}$

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Question

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

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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}$

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Question

Given , find the general form for the antiderivative .

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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 )!

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Question

Compute the following integral:

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

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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.

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Question

Solve:

$\int $(5\sec^2$(x)+\sec(x)\tan(x))dx$

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Answer

The integral can be solved knowing the derivatives of the following functions:

$$\frac{\mathrm{d}$ }{\mathrm{d} x} $\tan(x)=\sec^2$(x)$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$

Given that the integrand is simply the sum of these two derivatives, we find that our integral is equal to

$\int $(5\sec^2$(x)+\sec(x)\tan(x))dx=5\tan(x)+\sec(x)+C$

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