Integrals

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AP Calculus AB › Integrals

Questions 1 - 10

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

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

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

Integrate:

$\int 7x^4 \sin(3x^5)dx$

$-\frac{7}{15}\cos(3x^5)+C$

$\frac{7}{15}\cos(3x^5)+C$

$-\frac{7}{15}\cos(3x^5)$

$-\cos(3x^5)+C$

Explanation

To integrate, the following substitution was made:

Now, we rewrite the integral in terms of u and integrate:

$\frac{7}{15}\int\sin(u)du=-\frac{7}{15}\cos(u)+C$

The following rule was used for integration:

$\int \sin(x)dx=-\cos(x)+C$

Finally, rewrite the final answer in terms of our original x term:

$-\frac{7}{15}\sin(3x^5)+C$

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

Explanation

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

Given , find the general form for the antiderivative .

$\frac{4}{3}x^3 +10x^2 +25x+C$

None of the other answers

Explanation

To answer this, we will need to FOIL our function first.

Now can find the antiderivatives of each of these three summands using the power rule.

$f(x) = \frac{4}{3}x^3 +10x^2 +25x+C$ (Don't forget )!

Find the antiderivative of the following.

$\sec^2(x)$

$\tan (x)$

$2\sec (x)$

$\frac{\tan (x)}{2}$

$2\tan (x)$

Explanation

$\sec^2(x)$ is the derivative of $\tan (x)$ . Thus, the antiderivative of $\sec^2(x)$ is $\tan (x)$ .

$\begin{align*}&\text{Evaluate the integral: }\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx\end{align*}$

Explanation

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[cos(ax)]=\frac{sin(ax)}{a}\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=\frac{(27sin(x + 1))}{8}|_{8}^{11.5}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=\frac{(27sin((11.5) + 1))}{8}-(\frac{(27sin((8) + 1))}{8})\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=-1.61\end{align*}$

Integrate,

$\int[x^2 +\sin(x)-3]dx$

$\frac{1}{3}x^3 - \cos(x) - 3x +C$

$\frac{1}{2}x^2 + \cos(x) - 3x +C$

$x -x^2+ \cos(x) +C$

$\frac{1}{2}x - \sin(x) - 3x^2 +C$

$\frac{1}{2}x^3 -\cos(x) - \frac{3}{2}x^2 +C$

Explanation

Integrate

$\int[x^2 +\sin(x)-3]dx$

1) Apply the sum rule for integration,

$\int x^2 dx +\int \sin(x)dx -\int 3dx$

2) Integrate each individual term and include a constant of integration,

$\frac{1}{3}x^3 - \cos(x) - 3x +C$

Further Discussion

Since indefinite integration is essentially a reverse process of differentiation, check your result by computing its' derivative.

$\frac{d}{dx}\left(\frac{1}{3}x^3-\cos(x)-3x+C \right )=x^2+\sin(x) -3$

This is the same function we integrated, which confirms our result. Also, because the derivative of a constant is always zero, we must include "C" in our result since any constant added to any function will produce the same derivative.

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