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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #81 : Calculus Ii — Integrals

What is the indefinite integral of 4x+3 with respect to ?

Possible Answers:

2x^2+3x+c

4+c

2x^2+3x

2x^3-c

Correct answer:

2x^2+3x+c

Explanation:

To find the indefinite integral, we're going to use the reverse power rule: raise the exponent of the variable by one and then divide by that new exponent.

$\int 4x+3 dx=(\frac{4x^{1+1}}{1+1})+(\frac{3x^{0+1}}{0+1})+c$

Be sure to include + c to compensate for any constant!

$\int 4x+3dx=(\frac{4x^{1+1}}{1+1})+(\frac{3x^{0+1}}{0+1})+c$

$\int 4x+3dx=(\frac{4x^{2}}{2})+(\frac{3x^{1}}{1})+c$

$\int 4x+3dx=2x^2+3x+c$

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Example Question #2204 : High School Math

$\int_3^4x^3+2x^2+5=?$

Possible Answers:

$53\tfrac{1}{4}$

$179\tfrac{7}{12}$

$73\tfrac{5}{12}$

$126\tfrac{2}{3}$

Correct answer:

$73\tfrac{5}{12}$

Explanation:

The Fundamental Theorem of Calculus states that if g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ . Therefore, we need to find the indefinite integral of our given equation.

To find the indefinite integral, we can use the reverse power rule. Raise the exponent of the variable by one and then divide by that new exponent.

We're going to treat as 5x^0 .

$\int x^3+2x^2+5{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{2+1}}{2+1}+\frac{5x^{0+1}}{0+1}+c$

Remember to include the when taking the integral to compensate for any constant.

$\int x^3+2x^2+5{\mathrm{d} x}=\frac{x^{4}}{4}+\frac{2x^{3}}{3}+\frac{5x^{1}}{1}+c$

Simplify.

$\int x^3+2x^2+5{\mathrm{d} x}=\frac{1}{4}x^4+\frac{2}{3}x^3+5x+c$

Plug that into FTOC:

$\int_a^b x^3+2x^2+5{\mathrm{d} x}=(\frac{1}{4}(b)^4+\frac{2}{3}(b)^3+5(b)+c)-(\frac{1}{4}(a)^4+\frac{2}{3}(a)^3+5(a)+c)$

Notice that the 's cancel out.

Plug in our given numbers.

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(\frac{1}{4}(4)^4+\frac{2}{3}(4)^3+5(4))-(\frac{1}{4}(3)^4+\frac{2}{3}(3)^3+5(3))$ $\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(\frac{1}{4}(256)+\frac{2}{3}(64)+20)-(\frac{1}{4}(81)+\frac{2}{3}(27)+15)$

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(64+42\tfrac{2}{3}+20)-(20\tfrac{1}{4}+18+15)$

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(126\tfrac{2}{3})-(53\tfrac{1}{4})$

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=73\tfrac{5}{12}$

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Example Question #2205 : High School Math

$\int_2^44x^2+\frac{2}{3}x{\mathrm{d} x}=?$

Possible Answers:

$78\tfrac{1}{3}$

$102\tfrac{2}{3}$

308

$78\tfrac{2}{3}$

708

Correct answer:

$78\tfrac{2}{3}$

Explanation:

To find the definite integral, we can use the Fundamental Theorem of Calculus that states that if g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ .

Therefore, we need to find the indefinite integral of our equation to start.

To find the indefinite integral, we can use the reverse power rule. We raise the exponent of the variable by one and divide by our new exponent. For this problem that would look like this:

$\int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4x^{2+1}}{2+1}+\frac{\frac{2}{3}x^{1+1}}{1+1}+c$

Remember to include a to cover any potential constant that might be in our new equation.

$\int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4x^{3}}{3}+\frac{\frac{2}{3}x^{2}}{2}+c$

$\int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4}{3}x^3+\frac{1}{3}x^2+c$

Plug that into FTOC:

$\int_a^b 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}b^3+\frac{1}{3}b^2+c)-(\frac{4}{3}a^3+\frac{1}{3}a^2+c)$

Notice that the 's cancel out.

Plug in our given values.

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(4)^3+\frac{1}{3}(4)^2)-(\frac{4}{3}(2)^3+\frac{1}{3}(2)^2)$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(64)+\frac{1}{3}(16))-(\frac{4}{3}(8)+\frac{1}{3}(4))$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{256}{3}+\frac{16}{3})-(\frac{32}{3}+\frac{4}{3})$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{272}{3})-(\frac{36}{3})$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{236}{3}$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=78\tfrac{2}{3}$

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Example Question #2206 : High School Math

$\int_3^64x^2+\frac{2}{3}x{\mathrm{d} x}=?$

Possible Answers:

7.69

339

300

261

Correct answer:

261

Explanation:

To find the definite integral, we can use the Fundamental Theorem of Calculus which states that if g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ .

Therefore, we need to find the indefinite integral of our equation to start.

To find the indefinite integral, we can use the reverse power rule. We raise the exponent of the variable by one and divide by our new exponent.

$\int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4x^{2+1}}{2+1}+\frac{\frac{2}{3}x^{1+1}}{1+1}+c$

Remember to include a to cover any potential constant that might be in our new equation.

$\int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4x^{3}}{3}+\frac{\frac{2}{3}x^{2}}{2}+c$

$\int 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{4}{3}x^3+\frac{1}{3}x^2+c$

Plug that into FTOC:

$\int_a^b 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}b^3+\frac{1}{3}b^2+c)-(\frac{4}{3}a^3+\frac{1}{3}a^2+c)$

Notice that the 's cancel out.

Plug in our given values.

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(6)^3+\frac{1}{3}(6)^2)-(\frac{4}{3}(3)^3+\frac{1}{3}(3)^2)$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(216)+\frac{1}{3}(36))-(\frac{4}{3}(27)+\frac{1}{3}(9))$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(288+12)-(36+3)$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(300)-(39)$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=261$

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Example Question #21 : Finding Definite Integrals

$\int_{-\pi}^{\pi}\sin(x){\mathrm{d} x}=?$

Possible Answers:

$2\pi$

Correct answer:

Explanation:

The fundamental theorem of calculus states that if f'(x)=g(x) , then $\int_a^bg(x){\mathrm{d} x}=f(b)-f(a)$ .

First, we need to find the indefinite integral of our given equation. Just like with the derivatives, the indefinite integrals or anti-derivatives of trig functions must be memorized.

$\int \sin(x){\mathrm{d} x}=-\cos(x)+c$

Don't forget the to compensate for any potential constant!

Plug this in to our FTOC:

$\int_b^a \sin(x){\mathrm{d} x}=(-\cos(b)+c)-(-\cos(a)+c)$ .

Notice that the 's cancel out.

$\int_b^a \sin(x){\mathrm{d} x}=-\cos(b)+\cos(a)$ .

Now plug in the given values.

$\int_{-\pi}^{\pi} \sin(x){\mathrm{d} x}=-\cos(\pi)+\cos(-\pi)$

$\int_{-\pi}^{\pi} \sin(x){\mathrm{d} x}=0+0$

$\int_{-\pi}^{\pi} \sin(x){\mathrm{d} x}=0$

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Example Question #29 : Finding Integrals

$\int_2^8 4x^2+5x-3\: dx=?$

Possible Answers:

216

804

810

$818\tfrac{2}{3}$

270

Correct answer:

804

Explanation:

To solve for the definite integral, use the fundamental theorem of calculus. If g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ .

First we need to find the indefinite integral.

To find the indefinite integral of our given equation, we can use the reverse power rule: we raise the exponent by one and then divide by that new exponent.

$\int 4x^2+5x-3\: dx=\frac{4x^{2+1}}{2+1}+\frac{5x^{1+1}}{1+1}-\frac{3x^{0+1}}{0+1}+c$

Don't forget to include a to compensate for any constant!

$\int 4x^2+5x-3\: dx=\frac{4x^{3}}{3}+\frac{5x^{2}}{2}-\frac{3x^{1}}{1}+c$

$\int 4x^2+5x-3\: dx=\frac{4}{3}x^3+\frac{5}{2}x^2-3x+c$

Plug this into our first FTOC equation:

$\int_a^b 4x^2+5x-3\: dx=(\frac{4}{3}b^3+\frac{5}{2}b^2-3b+c)-(\frac{4}{3}a^3+\frac{5}{2}a^2-3a+c)$

Notice that the 's cancel out.

$\int_a^b 4x^2+5x-3\: dx=(\frac{4}{3}b^3+\frac{5}{2}b^2-3b)-(\frac{4}{3}a^3+\frac{5}{2}a^2-3a)$

Plug in our given values.

$\int_2^8 4x^2+5x-3\: dx=(\frac{4}{3}(8)^3+\frac{5}{2}(8)^2-3(8))-(\frac{4}{3}(2)^3+\frac{5}{2}(2)^2-3(2))$

$\int_2^8 4x^2+5x-3\: dx=(682\tfrac{2}{3}+160-24)-(10\tfrac{2}{3}+10-6)$

$\int_2^8 4x^2+5x-3\: dx=(818\tfrac{2}{3})-(14\tfrac{2}{3})$

$\int_2^8 4x^2+5x-3\: dx=804$

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Example Question #91 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Find the indefinite ingtegral for $\int\frac{x+1}{\sqrt{x}} dx$ .

Possible Answers:

$x^{}\frac{3}{2}+ x^{\frac{1}{2}}+C$

$(x-2)^\frac{3}{2}$

$\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C$

Correct answer:

$\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C$

Explanation:

First, bring up the radical into the numerator and distribute to the (x+1) term.

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

$\int x^{\frac{1}{2}} + x^{\frac{-1}{2}}dx$

Then integrate.

$\frac{2}{3}x^{\frac{3}{2}}+ 2x^{\frac{1}{2}}$

Since it's indefinite, don't forget to add the C:

$\frac{2}{3}x^{\frac{3}{2}}+ 2x^{\frac{1}{2}}+C$

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Example Question #111 : Asymptotic And Unbounded Behavior

Integrate this function: $\int(x+6)dx$ .

Possible Answers:

x^2+6x

$\frac{x^2}{2}+6x+C$

x^2 + 6x+C

Correct answer:

$\frac{x^2}{2}+6x+C$

Explanation:

First, divide up into two different integral expressions:

$\int xdx + \int6dx$

Then, integrate each:

$\frac{x^2}{2} + 6x$

Don't forget "C" because it is an indefinite integral:

$\frac{x^2}{2} + 6x+C$

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Example Question #93 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Integrate the following expression: $\int(2x^4-6x^3+9x)dx$ .

Possible Answers:

x^5-x^4+x^2+C

2x^5-3x^4+9x^2

$\frac{2x^5}{5}-\frac{3x^4}{2}+\frac{9x^2}{2}+C$

Correct answer:

$\frac{2x^5}{5}-\frac{3x^4}{2}+\frac{9x^2}{2}+C$

Explanation:

First, divide up into three different expressions so you can integrate each x term separately:

$\int2x^{4}dx-\int6x^{3}dx+\int9xdx$

Then, integrate and simplify:

$2(\frac{x^{5}}{5})-6(\frac{x^4}{4})+9(\frac{x^2}{2})$

$\frac{2x^5}{5}-\frac{3x^4}{2}+\frac{9x^2}{2}$

Don't forget "C" because it's an indefinite integral:

$\frac{2x^5}{5}-\frac{3x^4}{2}+\frac{9x^2}{2}+C$

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Example Question #94 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Find the general solution of $\int\frac{1}{x^3}dx$ to find the particular solution that satisfies the intitial condition F(1)=0

Possible Answers:

$F(x)=\frac{1}{2x^2}+\frac{1}{2}$

$F(x)=-\frac{1}{2x^2}+\frac{1}{2}$

$F(x)=-2x^2+\frac{1}{2}$

Correct answer:

$F(x)=-\frac{1}{2x^2}+\frac{1}{2}$

Explanation:

To start the problem, it's easier if you bring up the denominator and make it a negative exponent:

$\int x^{-3}dx$

Then, integrate:

$\frac{x^{-2}}{-2}$

Simplify and add the "C" for an indefinite integral:

$-\frac{1}{2x^2}+ C$

Plug in the initial conditions [F(1)=0] to find C and generate the particular solution:

$-\frac{1}{2x^2}+C=0$

$-\frac{1}{2(1^2)}+C=0$

$C=\frac{1}{2}$

Thus, your final equation is:

$F(x)=-\frac{1}{2x^2}+\frac{1}{2}$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #81 : Calculus Ii — Integrals

Example Question #2204 : High School Math

Example Question #2205 : High School Math

Example Question #2206 : High School Math

Example Question #21 : Finding Definite Integrals

Example Question #29 : Finding Integrals

Example Question #91 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #111 : Asymptotic And Unbounded Behavior

Example Question #93 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #94 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

All AP Calculus AB Resources

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