Integrals - AP Calculus AB

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Question

A ball is thrown into the air. It's height, after t seconds is modeled by the formula:

$h(t)=-15t^2$+30t feet.

At what time will the velocity equal zero?

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Answer

In order to find where the velocity is equal to zero, take the derivative of the function and set it equal to zero.

h(t) = –15t2 + 30t

h'(t) = –30t + 30

0 = –30t + 30

Then solve for "t".

–30 = –30t

t = 1

The velocity will be 0 at 1 second.

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Question

Evaluate the definite integral

$\int_${\pi}^{0}$$4(3x^2$-sin(x)+2)dx$

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Answer

Here we are using several basic properties of definite integrals as well as the fundamental theorem of calculus.

First, you can pull coefficients out to the front of integrals.

$\int_${\pi}^{0}$$4(3x^2$-sin(x)+2)dx=4\int_${\pi}^{0}$$(3x^2$-sin(x)+2)dx$

Second, we notice that our lower bound is bigger than our upper bound. You can switch the upper and lower bounds if you also switch the sign.

$4\int_${\pi}^{0}$$(3x^2$-sin(x)+2)dx=-4\int_${0}^{\pi}$$(3x^2$-sin(x)+2)dx$

Lastly, our integral "distributes" over addition and subtraction. That means you can split the integral by each term and integrate each term separately.

$-4\int_${0}^{\pi}$$(3x^2$-sin(x)+2)dx=-4(\int_${0}^{\pi}$$3x^2dx$-\int_${0}^{\pi}$sin(x)dx+\int_${0}^{\pi}$2dx)$

Now we integrate and calculate using the Fundamental Theorem of Calculus.

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Question

If are continuous functions, $g(x) = $\frac{1}{2}$f(x)$ , $\int_$0^4$ f(x) dx = 6$ , and $\int_$4^{10}$ g(x)dx = 8$ , find $\int_$0^{10}$ f(x) dx$ .

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Answer

We proceed as follows

$\int_$0^{10}$ f(x) dx$ . (Start)

$=\int_$0^4$ f(x) dx + \int_$4^{10}$f(x)dx$ . (Break up the integral using the additive rule.)

$=\int_$0^4$ f(x) dx + \int_$4^{10}$2g(x)dx$ . (We don't have information about the 2nd integral, so we solve our first equation for and replace it in this integral.)

$=\int_$0^4$ f(x) dx + 2\int_$4^{10}$g(x)dx$ . (Factor out the using linearity.)

. (Substitute in what we were given.)

.

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Question

$\begin{align}&\int_${24}^{9}$f(x)dx=-24\&\int_${24}^{26}$f(x)dx=11\&\int_${26}^{32}$f(x)dx=24\&\int_${32}^{34}$f(x)dx=34\&$\text{Calculate from the above values }$\int_${9}^{34}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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}^{34}$f(x)dx\&\int_${9}^{34}$f(x)dx=-(-24)+(11)+(24)+(34)\&\int_${9}^{34}$f(x)dx=93\&\end{align}$

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Question

$\begin{align}&\int_${13}^{2}$f(x)dx=-94\&\int_${13}^{19}$f(x)dx=-23\&\int_${23}^{19}$f(x)dx=-32\&\int_${28}^{23}$f(x)dx=-18\&\int_${31}^{28}$f(x)dx=-84\&$\text{Determine }$\int_${2}^{31}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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 }$\int_${2}^{31}$f(x)dx\&\int_${2}^{31}$f(x)dx=-\int_${13}^{2}$f(x)dx+\int_${13}^{19}$f(x)dx-\int_${23}^{19}$f(x)dx-\int_${28}^{23}$f(x)dx-\int_${31}^{28}$f(x)dx\&\int_${2}^{31}$f(x)dx=-(-94)+(-23)-(-32)-(-18)-(-84)\&\int_${2}^{31}$f(x)dx=205\&\end{align}$

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Question

$\begin{align}&\int_${-10}^{-5}$f(x)dx=95\&\int_${-1}^{-5}$f(x)dx=-41\&\int_${6}^{-1}$f(x)dx=-26\&\int_${21}^{6}$f(x)dx=-38\&\int_${28}^{21}$f(x)dx=62\&$\text{Calculate from the above values }$\int_${-10}^{28}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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 }$\int_${-10}^{28}$f(x)dx\&\int_${-10}^{28}$f(x)dx=\int_${-10}^{-5}$f(x)dx-\int_${-1}^{-5}$f(x)dx-\int_${6}^{-1}$f(x)dx-\int_${21}^{6}$f(x)dx-\int_${28}^{21}$f(x)dx\&\int_${-10}^{28}$f(x)dx=(95)-(-41)-(-26)-(-38)-(62)\&\int_${-10}^{28}$f(x)dx=138\&\end{align}$

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Question

$\begin{align}&\int_${-3}^{9}$f(x)dx=-67\&\int_${9}^{22}$f(x)dx=74\&\int_${37}^{22}$f(x)dx=-18\&\int_${39}^{37}$f(x)dx=-67\&\int_${42}^{39}$f(x)dx=68\&\int_${42}^{45}$f(x)dx=86\&$\text{Calculate from the above values }$\int_${-3}^{45}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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 }$\int_${-3}^{45}$f(x)dx\&\int_${-3}^{45}$f(x)dx=\int_${-3}^{9}$f(x)dx+\int_${9}^{22}$f(x)dx-\int_${37}^{22}$f(x)dx-\int_${39}^{37}$f(x)dx-\int_${42}^{39}$f(x)dx+\int_${42}^{45}$f(x)dx\&\int_${-3}^{45}$f(x)dx=(-67)+(74)-(-18)-(-67)-(68)+(86)\&\int_${-3}^{45}$f(x)dx=110\&\end{align}$

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Question

$\begin{align}&\int_${-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}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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}$f(x)dx=88\&\end{align}$

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Question

$\begin{align}&\int_${-10}^{3}$f(x)dx=95\&\int_${3}^{13}$f(x)dx=63\&\int_${23}^{13}$f(x)dx=78\&\int_${23}^{37}$f(x)dx=-54\&\int_${37}^{50}$f(x)dx=-56\&\int_${50}^{56}$f(x)dx=77\&$\text{Use the above values to find }$\int_${-10}^{56}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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 }$\int_${-10}^{56}$f(x)dx\&\int_${-10}^{56}$f(x)dx=\int_${-10}^{3}$f(x)dx+\int_${3}^{13}$f(x)dx-\int_${23}^{13}$f(x)dx+\int_${23}^{37}$f(x)dx+\int_${37}^{50}$f(x)dx+\int_${50}^{56}$f(x)dx\&\int_${-10}^{56}$f(x)dx=(95)+(63)-(78)+(-54)+(-56)+(77)\&\int_${-10}^{56}$f(x)dx=47\&\end{align}$

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Question

$\begin{align}&\int_${-4}^{6}$f(x)dx=5\&\int_${12}^{6}$f(x)dx=90\&\int_${15}^{12}$f(x)dx=73\&\int_${19}^{15}$f(x)dx=-70\&\int_${22}^{19}$f(x)dx=97\&\int_${24}^{22}$f(x)dx=36\&$\text{Calculate from the above values }$\int_${-4}^{24}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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 }$\int_${-4}^{24}$f(x)dx\&\int_${-4}^{24}$f(x)dx=\int_${-4}^{6}$f(x)dx-\int_${12}^{6}$f(x)dx-\int_${15}^{12}$f(x)dx-\int_${19}^{15}$f(x)dx-\int_${22}^{19}$f(x)dx-\int_${24}^{22}$f(x)dx\&\int_${-4}^{24}$f(x)dx=(5)-(90)-(73)-(-70)-(97)-(36)\&\int_${-4}^{24}$f(x)dx=-221\&\end{align}$

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Question

$\begin{align}&\int_${-4}^{0}$f(x)dx=-13\&\int_${15}^{0}$f(x)dx=56\&\int_${15}^{23}$f(x)dx=-53\&\int_${23}^{31}$f(x)dx=96\&\int_${31}^{32}$f(x)dx=-44\&$\text{Calculate from the above values }$\int_${-4}^{32}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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 }$\int_${-4}^{32}$f(x)dx\&\int_${-4}^{32}$f(x)dx=\int_${-4}^{0}$f(x)dx-\int_${15}^{0}$f(x)dx+\int_${15}^{23}$f(x)dx+\int_${23}^{31}$f(x)dx+\int_${31}^{32}$f(x)dx\&\int_${-4}^{32}$f(x)dx=(-13)-(56)+(-53)+(96)+(-44)\&\int_${-4}^{32}$f(x)dx=-70\&\end{align}$

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Question

$\begin{align}&\int_${3}^{6}$f(x)dx=-8\&\int_${6}^{16}$f(x)dx=-87\&\int_${24}^{16}$f(x)dx=-85\&\int_${39}^{24}$f(x)dx=48\&\int_${53}^{39}$f(x)dx=42\&$\text{Determine }$\int_${3}^{53}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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 }$\int_${3}^{53}$f(x)dx\&\int_${3}^{53}$f(x)dx=\int_${3}^{6}$f(x)dx+\int_${6}^{16}$f(x)dx-\int_${24}^{16}$f(x)dx-\int_${39}^{24}$f(x)dx-\int_${53}^{39}$f(x)dx\&\int_${3}^{53}$f(x)dx=(-8)+(-87)-(-85)-(48)-(42)\&\int_${3}^{53}$f(x)dx=-100\&\end{align}$

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Question

$\begin{align}&\int_${11}^{2}$f(x)dx=-56\&\int_${25}^{11}$f(x)dx=-52\&\int_${25}^{40}$f(x)dx=-82\&\int_${40}^{46}$f(x)dx=-71\&\int_${59}^{46}$f(x)dx=-23\&$\text{Calculate from the above values }$\int_${2}^{59}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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 }$\int_${2}^{59}$f(x)dx\&\int_${2}^{59}$f(x)dx=-\int_${11}^{2}$f(x)dx-\int_${25}^{11}$f(x)dx+\int_${25}^{40}$f(x)dx+\int_${40}^{46}$f(x)dx-\int_${59}^{46}$f(x)dx\&\int_${2}^{59}$f(x)dx=-(-56)-(-52)+(-82)+(-71)-(-23)\&\int_${2}^{59}$f(x)dx=-22\&\end{align}$

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Question

$\begin{align}&\int_${3}^{10}$f(x)dx=38\&\int_${25}^{10}$f(x)dx=-14\&\int_${30}^{25}$f(x)dx=76\&\int_${30}^{36}$f(x)dx=27\&\int_${36}^{45}$f(x)dx=9\&$\text{Use the above values to find }$\int_${3}^{45}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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 }$\int_${3}^{45}$f(x)dx\&\int_${3}^{45}$f(x)dx=\int_${3}^{10}$f(x)dx-\int_${25}^{10}$f(x)dx-\int_${30}^{25}$f(x)dx+\int_${30}^{36}$f(x)dx+\int_${36}^{45}$f(x)dx\&\int_${3}^{45}$f(x)dx=(38)-(-14)-(76)+(27)+(9)\&\int_${3}^{45}$f(x)dx=12\&\end{align}$

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Question

$\begin{align}&\int_${22}^{8}$f(x)dx=-68\&\int_${22}^{28}$f(x)dx=68\&\int_${28}^{29}$f(x)dx=44\&$\text{Calculate from the above values }$\int_${8}^{29}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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 }$\int_${8}^{29}$f(x)dx\&\int_${8}^{29}$f(x)dx=-\int_${22}^{8}$f(x)dx+\int_${22}^{28}$f(x)dx+\int_${28}^{29}$f(x)dx\&\int_${8}^{29}$f(x)dx=-(-68)+(68)+(44)\&\int_${8}^{29}$f(x)dx=180\&\end{align}$

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Question

$\begin{align}&\int_${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}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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}$f(x)dx=-40\&\end{align}$

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Question

$\begin{align}&\int_${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}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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}$f(x)dx=30\&\end{align}$

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Question

$\begin{align}&\int_${0}^{13}$f(x)dx=-60\&\int_${13}^{27}$f(x)dx=-25\&\int_${45}^{34}$f(x)dx=-22\&\int_${53}^{45}$f(x)dx=16\&\int_${53}^{54}$f(x)dx=-26\&\int_${0}^{54}$f(x)dx=-86\&$\text{Use the above values to find }$\int_${27}^{34}$f(x)dx\end{align}$

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Answer

$\begin{align}&$\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}$\end{align}$

$\begin{align}\&$\text{will give a negative of the original value:}$\&\int_$a^b$=-\int_$b^a$\&$\text{Knowing all this, we can calculate }$\int_${27}^{34}$f(x)dx\&\int_${0}^{54}$f(x)dx=\int_${0}^{13}$f(x)dx+\int_${13}^{27}$f(x)dx+\int_${27}^{34}$f(x)dx-\int_${45}^{34}$f(x)dx-\int_${53}^{45}$f(x)dx+\int_${53}^{54}$f(x)dx\&\int_${27}^{34}$f(x)dx=\int_${0}^{54}$f(x)dx-\int_${0}^{13}$f(x)dx-\int_${13}^{27}$f(x)dx+\int_${45}^{34}$f(x)dx+\int_${53}^{45}$f(x)dx-\int_${53}^{54}$f(x)dx\&\int_${27}^{34}$f(x)dx=-86-(-60)-(-25)+(-22)+(16)-(-26)\&\int_${27}^{34}$f(x)dx=19\&\end{align}$

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Question

$\begin{align}&\int_${10}^{15}$f(x)dx=-100\&\int_${15}^{29}$f(x)dx=11\&\int_${8}^{29}$f(x)dx=-184\&$\text{Calculate from the above values }$\int_${8}^{10}$f(x)dx\end{align}$

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$\begin{align}&$\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 }$\int_${8}^{10}$f(x)dx\&\int_${8}^{29}$f(x)dx=\int_${8}^{10}$f(x)dx+\int_${10}^{15}$f(x)dx+\int_${15}^{29}$f(x)dx\&\int_${8}^{10}$f(x)dx=\int_${8}^{29}$f(x)dx-\int_${10}^{15}$f(x)dx-\int_${15}^{29}$f(x)dx\&\int_${8}^{10}$f(x)dx=-184-(-100)-(11)\&\int_${8}^{10}$f(x)dx=-95\&\end{align}$

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Question

$\begin{align}&\int_${-5}^{8}$f(x)dx=68\&\int_${10}^{8}$f(x)dx=96\&\int_${10}^{22}$f(x)dx=24\&\int_${-10}^{22}$f(x)dx=-22\&$\text{Calculate from the above values }$\int_${-10}^{-5}$f(x)dx\end{align}$

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$\begin{align}&$\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}^{-5}$f(x)dx\&\int_${-10}^{22}$f(x)dx=\int_${-10}^{-5}$f(x)dx+\int_${-5}^{8}$f(x)dx-\int_${10}^{8}$f(x)dx+\int_${10}^{22}$f(x)dx\&\int_${-10}^{-5}$f(x)dx=\int_${-10}^{22}$f(x)dx-\int_${-5}^{8}$f(x)dx+\int_${10}^{8}$f(x)dx-\int_${10}^{22}$f(x)dx\&\int_${-10}^{-5}$f(x)dx=-22-(68)+(96)-(24)\&\int_${-10}^{-5}$f(x)dx=-18\&\end{align}$

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