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AP Calculus AB : Interpretations and properties of definite integrals

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Example Questions

Example Question #11 : Basic Properties Of Definite Integrals (Additivity And Linearity)

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

Possible Answers:

$\displaystyle 30$

$\displaystyle 48$

$\displaystyle -30$

$\displaystyle -48$

Correct answer:

$\displaystyle 30$

Explanation:

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

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Example Question #41 : Interpretations And Properties Of Definite Integrals

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

Possible Answers:

$\displaystyle -19$

$\displaystyle 31$

$\displaystyle -31$

$\displaystyle 19$

Correct answer:

$\displaystyle 19$

Explanation:

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

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Example Question #41 : Interpretations And Properties Of Definite Integrals

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

Possible Answers:

$\displaystyle -95$

$\displaystyle -82$

$\displaystyle 95$

$\displaystyle 82$

Correct answer:

$\displaystyle -95$

Explanation:

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

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Example Question #43 : Interpretations And Properties Of Definite Integrals

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

Possible Answers:

$\displaystyle 18$

$\displaystyle -210$

$\displaystyle -18$

210 $\displaystyle 210$

Correct answer:

$\displaystyle -18$

Explanation:

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

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Example Question #44 : Interpretations And Properties Of Definite Integrals

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

Possible Answers:

$\displaystyle 19$

$\displaystyle -19$

$\displaystyle 15$

$\displaystyle -15$

Correct answer:

$\displaystyle 15$

Explanation:

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

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Example Question #41 : Interpretations And Properties Of Definite Integrals

$\begin{align*}&\int_{-3}^{3}f(x)dx=95\\&\int_{15}^{3}f(x)dx=-81\\&\int_{-3}^{26}f(x)dx=154\\&\text{Use the above values to find }\int_{26}^{15}f(x)dx\end{align*}$ $\displaystyle \begin{align*}&\int_{-3}^{3}f(x)dx=95\\&\int_{15}^{3}f(x)dx=-81\\&\int_{-3}^{26}f(x)dx=154\\&\text{Use the above values to find }\int_{26}^{15}f(x)dx\end{align*}$

Possible Answers:

$\displaystyle 22$

$\displaystyle -140$

$\displaystyle -22$

140 $\displaystyle 140$

Correct answer:

$\displaystyle 22$

Explanation:

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

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Example Question #21 : Basic Properties Of Definite Integrals (Additivity And Linearity)

$\begin{align*}&\int_{-4}^{-3}f(x)dx=66\\&\int_{-3}^{4}f(x)dx=-32\\&\int_{6}^{14}f(x)dx=25\\&\int_{-4}^{14}f(x)dx=3\\&\text{Calculate from the above values }\int_{6}^{4}f(x)dx\end{align*}$ $\displaystyle \begin{align*}&\int_{-4}^{-3}f(x)dx=66\\&\int_{-3}^{4}f(x)dx=-32\\&\int_{6}^{14}f(x)dx=25\\&\int_{-4}^{14}f(x)dx=3\\&\text{Calculate from the above values }\int_{6}^{4}f(x)dx\end{align*}$

Possible Answers:

$\displaystyle 56$

$\displaystyle -40$

$\displaystyle 40$

$\displaystyle -56$

Correct answer:

$\displaystyle 56$

Explanation:

$\begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\\&\text{will give a negative of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing all this, we can calculate }\int_{6}^{4}f(x)dx\\&\int_{-4}^{14}f(x)dx=\int_{-4}^{-3}f(x)dx+\int_{-3}^{4}f(x)dx-\int_{6}^{4}f(x)dx+\int_{6}^{14}f(x)dx\\&-\int_{6}^{4}f(x)dx=\int_{-4}^{14}f(x)dx-\int_{-4}^{-3}f(x)dx-\int_{-3}^{4}f(x)dx-\int_{6}^{14}f(x)dx\\&-\int_{6}^{4}f(x)dx=3-(66)-(-32)-(25)\\&\int_{6}^{4}f(x)dx=56\\&\end{align*}$ $\displaystyle \begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\\&\text{will give a negative of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing all this, we can calculate }\int_{6}^{4}f(x)dx\\&\int_{-4}^{14}f(x)dx=\int_{-4}^{-3}f(x)dx+\int_{-3}^{4}f(x)dx-\int_{6}^{4}f(x)dx+\int_{6}^{14}f(x)dx\\&-\int_{6}^{4}f(x)dx=\int_{-4}^{14}f(x)dx-\int_{-4}^{-3}f(x)dx-\int_{-3}^{4}f(x)dx-\int_{6}^{14}f(x)dx\\&-\int_{6}^{4}f(x)dx=3-(66)-(-32)-(25)\\&\int_{6}^{4}f(x)dx=56\\&\end{align*}$

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Example Question #22 : Basic Properties Of Definite Integrals (Additivity And Linearity)

$\begin{align*}&\int_{10}^{19}f(x)dx=-40\\&\int_{19}^{30}f(x)dx=1\\&\int_{42}^{30}f(x)dx=-12\\&\int_{4}^{42}f(x)dx=-19\\&\text{Use the above values to find }\int_{10}^{4}f(x)dx\end{align*}$ $\displaystyle \begin{align*}&\int_{10}^{19}f(x)dx=-40\\&\int_{19}^{30}f(x)dx=1\\&\int_{42}^{30}f(x)dx=-12\\&\int_{4}^{42}f(x)dx=-19\\&\text{Use the above values to find }\int_{10}^{4}f(x)dx\end{align*}$

Possible Answers:

$\displaystyle 32$

$\displaystyle -32$

$\displaystyle -8$

$\displaystyle 8$

Correct answer:

$\displaystyle -8$

Explanation:

$\begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\\&\text{will give a negative of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing all this, we can calculate }\int_{10}^{4}f(x)dx\\&\int_{4}^{42}f(x)dx=-\int_{10}^{4}f(x)dx+\int_{10}^{19}f(x)dx+\int_{19}^{30}f(x)dx-\int_{42}^{30}f(x)dx\\&-\int_{10}^{4}f(x)dx=\int_{4}^{42}f(x)dx-\int_{10}^{19}f(x)dx-\int_{19}^{30}f(x)dx+\int_{42}^{30}f(x)dx\\&-\int_{10}^{4}f(x)dx=-19-(-40)-(1)+(-12)\\&\int_{10}^{4}f(x)dx=-8\\&\end{align*}$ $\displaystyle \begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\\&\text{will give a negative of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing all this, we can calculate }\int_{10}^{4}f(x)dx\\&\int_{4}^{42}f(x)dx=-\int_{10}^{4}f(x)dx+\int_{10}^{19}f(x)dx+\int_{19}^{30}f(x)dx-\int_{42}^{30}f(x)dx\\&-\int_{10}^{4}f(x)dx=\int_{4}^{42}f(x)dx-\int_{10}^{19}f(x)dx-\int_{19}^{30}f(x)dx+\int_{42}^{30}f(x)dx\\&-\int_{10}^{4}f(x)dx=-19-(-40)-(1)+(-12)\\&\int_{10}^{4}f(x)dx=-8\\&\end{align*}$

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Example Question #23 : Basic Properties Of Definite Integrals (Additivity And Linearity)

$\begin{align*}&\int_{-4}^{1}f(x)dx=-81\\&\int_{14}^{10}f(x)dx=80\\&\int_{14}^{24}f(x)dx=-4\\&\int_{24}^{31}f(x)dx=96\\&\int_{-4}^{31}f(x)dx=-11\\&\text{Use the above values to find }\int_{10}^{1}f(x)dx\end{align*}$ $\displaystyle \begin{align*}&\int_{-4}^{1}f(x)dx=-81\\&\int_{14}^{10}f(x)dx=80\\&\int_{14}^{24}f(x)dx=-4\\&\int_{24}^{31}f(x)dx=96\\&\int_{-4}^{31}f(x)dx=-11\\&\text{Use the above values to find }\int_{10}^{1}f(x)dx\end{align*}$

Possible Answers:

102 $\displaystyle 102$

$\displaystyle 58$

$\displaystyle -58$

$\displaystyle -102$

Correct answer:

$\displaystyle -58$

Explanation:

$\begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\\&\text{will give a negative of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing all this, we can calculate }\int_{10}^{1}f(x)dx\\&\int_{-4}^{31}f(x)dx=\int_{-4}^{1}f(x)dx-\int_{10}^{1}f(x)dx-\int_{14}^{10}f(x)dx+\int_{14}^{24}f(x)dx+\int_{24}^{31}f(x)dx\\&-\int_{10}^{1}f(x)dx=\int_{-4}^{31}f(x)dx-\int_{-4}^{1}f(x)dx+\int_{14}^{10}f(x)dx-\int_{14}^{24}f(x)dx-\int_{24}^{31}f(x)dx\\&-\int_{10}^{1}f(x)dx=-11-(-81)+(80)-(-4)-(96)\\&\int_{10}^{1}f(x)dx=-58\\&\end{align*}$ $\displaystyle \begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\\&\text{will give a negative of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing all this, we can calculate }\int_{10}^{1}f(x)dx\\&\int_{-4}^{31}f(x)dx=\int_{-4}^{1}f(x)dx-\int_{10}^{1}f(x)dx-\int_{14}^{10}f(x)dx+\int_{14}^{24}f(x)dx+\int_{24}^{31}f(x)dx\\&-\int_{10}^{1}f(x)dx=\int_{-4}^{31}f(x)dx-\int_{-4}^{1}f(x)dx+\int_{14}^{10}f(x)dx-\int_{14}^{24}f(x)dx-\int_{24}^{31}f(x)dx\\&-\int_{10}^{1}f(x)dx=-11-(-81)+(80)-(-4)-(96)\\&\int_{10}^{1}f(x)dx=-58\\&\end{align*}$

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Example Question #24 : Basic Properties Of Definite Integrals (Additivity And Linearity)

$\begin{align*}&\int_{0}^{4}f(x)dx=67\\&\int_{5}^{4}f(x)dx=-6\\&\int_{8}^{22}f(x)dx=72\\&\int_{22}^{23}f(x)dx=-54\\&\int_{0}^{23}f(x)dx=96\\&\text{Calculate from the above values }\int_{8}^{5}f(x)dx\end{align*}$ $\displaystyle \begin{align*}&\int_{0}^{4}f(x)dx=67\\&\int_{5}^{4}f(x)dx=-6\\&\int_{8}^{22}f(x)dx=72\\&\int_{22}^{23}f(x)dx=-54\\&\int_{0}^{23}f(x)dx=96\\&\text{Calculate from the above values }\int_{8}^{5}f(x)dx\end{align*}$

Possible Answers:

$\displaystyle -5$

$\displaystyle -17$

$\displaystyle 5$

$\displaystyle 17$

Correct answer:

$\displaystyle -5$

Explanation:

$\begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\\&\text{will give a negative of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing all this, we can calculate }\int_{8}^{5}f(x)dx\\&\int_{0}^{23}f(x)dx=\int_{0}^{4}f(x)dx-\int_{5}^{4}f(x)dx-\int_{8}^{5}f(x)dx+\int_{8}^{22}f(x)dx+\int_{22}^{23}f(x)dx\\&-\int_{8}^{5}f(x)dx=\int_{0}^{23}f(x)dx-\int_{0}^{4}f(x)dx+\int_{5}^{4}f(x)dx-\int_{8}^{22}f(x)dx-\int_{22}^{23}f(x)dx\\&-\int_{8}^{5}f(x)dx=96-(67)+(-6)-(72)-(-54)\\&\int_{8}^{5}f(x)dx=-5\\&\end{align*}$ $\displaystyle \begin{align*}&\text{A principle of integrals is one of summation. If the values}\\&\text{of an integral across given sets of points is known, they could}\\&\text{ be used to find integral values across other points. Imagine}\\&\text{a set of points that are in ascending numerical value: a, b,}\\&\text{c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\text{ and likewise: }\int_a^d-\int_a^b-\int_b^c=\int_c^d\\&\text{Also, if an integral value is known, the same integral backwards}\\&\text{will give a negative of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing all this, we can calculate }\int_{8}^{5}f(x)dx\\&\int_{0}^{23}f(x)dx=\int_{0}^{4}f(x)dx-\int_{5}^{4}f(x)dx-\int_{8}^{5}f(x)dx+\int_{8}^{22}f(x)dx+\int_{22}^{23}f(x)dx\\&-\int_{8}^{5}f(x)dx=\int_{0}^{23}f(x)dx-\int_{0}^{4}f(x)dx+\int_{5}^{4}f(x)dx-\int_{8}^{22}f(x)dx-\int_{22}^{23}f(x)dx\\&-\int_{8}^{5}f(x)dx=96-(67)+(-6)-(72)-(-54)\\&\int_{8}^{5}f(x)dx=-5\\&\end{align*}$

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AP Calculus AB : Interpretations and properties of definite integrals

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Example Question #11 : Basic Properties Of Definite Integrals (Additivity And Linearity)

Example Question #41 : Interpretations And Properties Of Definite Integrals

Example Question #41 : Interpretations And Properties Of Definite Integrals

Example Question #43 : Interpretations And Properties Of Definite Integrals

Example Question #44 : Interpretations And Properties Of Definite Integrals

Example Question #41 : Interpretations And Properties Of Definite Integrals

Example Question #21 : Basic Properties Of Definite Integrals (Additivity And Linearity)

Example Question #22 : Basic Properties Of Definite Integrals (Additivity And Linearity)

Example Question #23 : Basic Properties Of Definite Integrals (Additivity And Linearity)

Example Question #24 : Basic Properties Of Definite Integrals (Additivity And Linearity)

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