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

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

Explanation

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

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

Explanation

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

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

Explanation

$\begin{align*}&\text{One of the principles of integrals is one of summation. If the}\&\text{values of an integral across certain sets of points is known,}\&\text{this could potentially be used to find integral values at other}\&\text{points. Imagine a set of points that are in ascending numerical}\&\text{value: a, b, c, and d. It follows that:}\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\&\text{Another principle of integrals is that if an integral value}\&\text{is known, to take the same integral backwards will give a negative}\&\text{of the original value:}\&\int_a^b=-\int_b^a\&\text{Knowing this, we can calculate }\int_{-9}^{18}f(x)dx\&\int_{-9}^{18}f(x)dx=\int_{-9}^{0}f(x)dx-\int_{13}^{0}f(x)dx-\int_{18}^{13}f(x)dx\&\int_{-9}^{18}f(x)dx=(62)-(-46)-(20)\&\int_{-9}^{18}f(x)dx=88\&\end{align*}$

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

Explanation

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

Given that $\int_{5}^{1}(2u)dx = 8$ and $\int_{3}^{2}(-u)dx = 4$ , find the value of the following expression:

$\int_{1}^{3}(u)dx - \int_{5}^{4}(u)dx + \int_{2}^{4}(u)dx$

Explanation

First, simplifying the given's gives us

$\int_{5}^{1}(2u)dx = -2\int_{1}^{5}(u)dx = 8$

$\int_{1}^{5}(u)dx = -4$

And

$\int_{3}^{2}(-u)dx = \int_{2}^{3}(u)dx = 4$

Our goal is to get the given expression into terms of these two integrals. Our first step will be to try and get a $\int_{1}^{5}(u)dx$ from our expression.

First note,

$-\int_{5}^{4}(u)dx = \int_{4}^{5}(u)dx$

And for the third term,

$\int_{2}^{4}(u)dx = \int_{2}^{3}(u)dx + \int_{3}^{4}(u)dx$

Putting these facts together, we can rewrite the original expression as

$\int_{1}^{3}(u)dx +(\int_{4}^{5}(u)dx) + (\int_{2}^{3}(u)dx + \int_{3}^{4}(u)dx)$

Rearranging,

$(\int_{1}^{3}(u)dx + \int_{3}^{4}(u)dx + \int_{4}^{5}(u)dx) + \int_{2}^{3}(u)dx$

The three terms in parentheses can all be brought together, as the top limit of the previous integral matches the bottom limit of the next integral. Thus, we now have

$\int_{1}^{5}(u)dx + \int_{2}^{3}(u)dx$

Substituting in our given's, this simplifies to

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

Explanation

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

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

Explanation

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

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

Explanation

$\begin{align*}&\text{One of the principles of integrals is one of summation. If the}\&\text{values of an integral across certain sets of points is known,}\&\text{this could potentially be used to find integral values at other}\&\text{points. Imagine a set of points that are in ascending numerical}\&\text{value: a, b, c, and d. It follows that:}\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\&\text{Another principle of integrals is that if an integral value}\&\text{is known, to take the same integral backwards will give a negative}\&\text{of the original value:}\&\int_a^b=-\int_b^a\&\text{Knowing this, we can calculate }\int_{-4}^{24}f(x)dx\&\int_{-4}^{24}f(x)dx=\int_{-4}^{6}f(x)dx-\int_{12}^{6}f(x)dx-\int_{15}^{12}f(x)dx-\int_{19}^{15}f(x)dx-\int_{22}^{19}f(x)dx-\int_{24}^{22}f(x)dx\&\int_{-4}^{24}f(x)dx=(5)-(90)-(73)-(-70)-(97)-(36)\&\int_{-4}^{24}f(x)dx=-221\&\end{align*}$

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

Explanation

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

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

Explanation

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

Basic properties of definite integrals (additivity and linearity)

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation