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

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Question

Evaluate the definite integral

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

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.

$-4(x^3+cos(x)+2x)]_{0}^{\pi}=-4[(\pi^3+cos(\pi)+2\pi)-(0^3+cos(0)+2(0))]$

$=-4[(\pi^3-1+2\pi)-1]=-4\pi^3-8\pi+8$

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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 $\int_0^{10} f(x) dx$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer

$\begin{align}&\text{A principle of integrals is one of summation. If the values}\&\text{of an integral across given sets of points is known, they could}\&\text{ be used to find integral values across other points. Imagine}\&\text{a set of points that are in ascending numerical value: a, b,}\&\text{c, and d. It follows that:}\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\&\text{Also, if an integral value is known, the same integral backwards}\&\text{will give a negative of the original value:}\&\int_a^b=-\int_b^a\&\text{Knowing all this, we can calculate }\int_{8}^{10}f(x)dx\&\int_{8}^{29}f(x)dx=\int_{8}^{10}f(x)dx+\int_{10}^{15}f(x)dx+\int_{15}^{29}f(x)dx\&\int_{8}^{10}f(x)dx=\int_{8}^{29}f(x)dx-\int_{10}^{15}f(x)dx-\int_{15}^{29}f(x)dx\&\int_{8}^{10}f(x)dx=-184-(-100)-(11)\&\int_{8}^{10}f(x)dx=-95\&\end{align}$

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Question

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

Answer

$\begin{align}&\text{A principle of integrals is one of summation. If the values}\&\text{of an integral across given sets of points is known, they could}\&\text{ be used to find integral values across other points. Imagine}\&\text{a set of points that are in ascending numerical value: a, b,}\&\text{c, and d. It follows that:}\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\&\text{Also, if an integral value is known, the same integral backwards}\&\text{will give a negative of the original value:}\&\int_a^b=-\int_b^a\&\text{Knowing all this, we can calculate }\int_{-10}^{-5}f(x)dx\&\int_{-10}^{22}f(x)dx=\int_{-10}^{-5}f(x)dx+\int_{-5}^{8}f(x)dx-\int_{10}^{8}f(x)dx+\int_{10}^{22}f(x)dx\&\int_{-10}^{-5}f(x)dx=\int_{-10}^{22}f(x)dx-\int_{-5}^{8}f(x)dx+\int_{10}^{8}f(x)dx-\int_{10}^{22}f(x)dx\&\int_{-10}^{-5}f(x)dx=-22-(68)+(96)-(24)\&\int_{-10}^{-5}f(x)dx=-18\&\end{align}$

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Question

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

Answer

$\begin{align}&\text{A principle of integrals is one of summation. If the values}\&\text{of an integral across given sets of points is known, they could}\&\text{ be used to find integral values across other points. Imagine}\&\text{a set of points that are in ascending numerical value: a, b,}\&\text{c, and d. It follows that:}\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\&\text{Also, if an integral value is known, the same integral backwards}\&\text{will give a negative of the original value:}\&\int_a^b=-\int_b^a\&\text{Knowing all this, we can calculate }\int_{2}^{1}f(x)dx\&\int_{-2}^{24}f(x)dx=-\int_{1}^{-2}f(x)dx-\int_{2}^{1}f(x)dx-\int_{13}^{2}f(x)dx-\int_{24}^{13}f(x)dx\&-\int_{2}^{1}f(x)dx=\int_{-2}^{24}f(x)dx+\int_{1}^{-2}f(x)dx+\int_{13}^{2}f(x)dx+\int_{24}^{13}f(x)dx\&-\int_{2}^{1}f(x)dx=-17+(-65)+(86)+(-19)\&\int_{2}^{1}f(x)dx=15\&\end{align}$

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