Basic properties of definite integrals (additivity and linearity) - AP Calculus AB

Card 1 of 308

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

Evaluate the definite integral

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

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$ .

Tap to reveal answer

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.)

.

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=93\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=205\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=138\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=110\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=47\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=-221\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=-70\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=-100\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=-22\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=12\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=180\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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}$f(x)dx=19\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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 }$\int_${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}$f(x)dx=-95\&\end{align}$

← Didn't Know|Knew It →

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}$f(x)dx\end{align}$

Tap to reveal answer

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 }$\int_${-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}$f(x)dx=-18\&\end{align}$

← Didn't Know|Knew It →

Question

$\begin{align}&\int_${1}^{-2}$f(x)dx=-65\&\int_${13}^{2}$f(x)dx=86\&\int_${24}^{13}$f(x)dx=-19\&\int_${-2}^{24}$f(x)dx=-17\&$\text{Determine }$\int_${2}^{1}$f(x)dx\end{align}$

Tap to reveal answer

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 }$\int_${2}^{1}$f(x)dx\&\int_${-2}^{24}$f(x)dx=-\int_${1}^{-2}$f(x)dx-\int_${2}^{1}$f(x)dx-\int_${13}^{2}$f(x)dx-\int_${24}^{13}$f(x)dx\&-\int_${2}^{1}$f(x)dx=\int_${-2}^{24}$f(x)dx+\int_${1}^{-2}$f(x)dx+\int_${13}^{2}$f(x)dx+\int_${24}^{13}$f(x)dx\&-\int_${2}^{1}$f(x)dx=-17+(-65)+(86)+(-19)\&\int_${2}^{1}$f(x)dx=15\&\end{align}$

← Didn't Know|Knew It →

0

Knew It