\(\displaystyle \begin{align*}&\text{One of the principles of integrals is one of summation. If the}\\&\text{values of an integral across certain sets of points is known,}\\&\text{this could potentially be used to find integral values at other}\\&\text{points. Imagine a set of points that are in ascending numerical}\\&\text{value: a, b, c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\\&\text{Another principle of integrals is that if an integral value}\\&\text{is known, to take the same integral backwards will give a negative}\\&\text{of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing this, we can calculate }\int_{2}^{59}f(x)dx\\&\int_{2}^{59}f(x)dx=-\int_{11}^{2}f(x)dx-\int_{25}^{11}f(x)dx+\int_{25}^{40}f(x)dx+\int_{40}^{46}f(x)dx-\int_{59}^{46}f(x)dx\\&\int_{2}^{59}f(x)dx=-(-56)-(-52)+(-82)+(-71)-(-23)\\&\int_{2}^{59}f(x)dx=-22\\&\end{align*}\)

\(\displaystyle \begin{align*}&\text{One of the principles of integrals is one of summation. If the}\\&\text{values of an integral across certain sets of points is known,}\\&\text{this could potentially be used to find integral values at other}\\&\text{points. Imagine a set of points that are in ascending numerical}\\&\text{value: a, b, c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\\&\text{Another principle of integrals is that if an integral value}\\&\text{is known, to take the same integral backwards will give a negative}\\&\text{of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing this, we can calculate }\int_{3}^{45}f(x)dx\\&\int_{3}^{45}f(x)dx=\int_{3}^{10}f(x)dx-\int_{25}^{10}f(x)dx-\int_{30}^{25}f(x)dx+\int_{30}^{36}f(x)dx+\int_{36}^{45}f(x)dx\\&\int_{3}^{45}f(x)dx=(38)-(-14)-(76)+(27)+(9)\\&\int_{3}^{45}f(x)dx=12\\&\end{align*}\)

\(\displaystyle \begin{align*}&\text{One of the principles of integrals is one of summation. If the}\\&\text{values of an integral across certain sets of points is known,}\\&\text{this could potentially be used to find integral values at other}\\&\text{points. Imagine a set of points that are in ascending numerical}\\&\text{value: a, b, c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\\&\text{Another principle of integrals is that if an integral value}\\&\text{is known, to take the same integral backwards will give a negative}\\&\text{of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing this, we can calculate }\int_{8}^{29}f(x)dx\\&\int_{8}^{29}f(x)dx=-\int_{22}^{8}f(x)dx+\int_{22}^{28}f(x)dx+\int_{28}^{29}f(x)dx\\&\int_{8}^{29}f(x)dx=-(-68)+(68)+(44)\\&\int_{8}^{29}f(x)dx=180\\&\end{align*}\)

\(\displaystyle \begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\\&\text{will give a negative of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing all this, we can calculate }\int_{48}^{33}f(x)dx\\&\int_{4}^{48}f(x)dx=-\int_{13}^{4}f(x)dx-\int_{17}^{13}f(x)dx+\int_{17}^{28}f(x)dx+\int_{28}^{33}f(x)dx-\int_{48}^{33}f(x)dx\\&-\int_{48}^{33}f(x)dx=\int_{4}^{48}f(x)dx+\int_{13}^{4}f(x)dx+\int_{17}^{13}f(x)dx-\int_{17}^{28}f(x)dx-\int_{28}^{33}f(x)dx\\&-\int_{48}^{33}f(x)dx=-10+(42)+(-56)-(35)-(-99)\\&\int_{48}^{33}f(x)dx=-40\\&\end{align*}\)

\(\displaystyle \begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\\&\text{will give a negative of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing all this, we can calculate }\int_{5}^{0}f(x)dx\\&\int_{0}^{26}f(x)dx=-\int_{5}^{0}f(x)dx+\int_{5}^{12}f(x)dx-\int_{26}^{12}f(x)dx\\&-\int_{5}^{0}f(x)dx=\int_{0}^{26}f(x)dx-\int_{5}^{12}f(x)dx+\int_{26}^{12}f(x)dx\\&-\int_{5}^{0}f(x)dx=108-(99)+(-39)\\&\int_{5}^{0}f(x)dx=30\\&\end{align*}\)

\(\displaystyle \begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\end{align*}\)

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

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

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

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

\(\displaystyle \begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\\&\text{will give a negative of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing all this, we can calculate }\int_{26}^{15}f(x)dx\\&\int_{-3}^{26}f(x)dx=\int_{-3}^{3}f(x)dx-\int_{15}^{3}f(x)dx-\int_{26}^{15}f(x)dx\\&-\int_{26}^{15}f(x)dx=\int_{-3}^{26}f(x)dx-\int_{-3}^{3}f(x)dx+\int_{15}^{3}f(x)dx\\&-\int_{26}^{15}f(x)dx=154-(95)+(-81)\\&\int_{26}^{15}f(x)dx=22\\&\end{align*}\)