Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #1201 : Functions

\(\displaystyle \int_{0}^{2}3x^3-2x^2+1dx\)

Possible Answers:

\(\displaystyle 3\)

\(\displaystyle \frac{26}{3}\)

\(\displaystyle 26\)

\(\displaystyle \frac{1}{26}\)

\(\displaystyle -\frac{26}{3}\)

Correct answer:

\(\displaystyle \frac{26}{3}\)

Explanation:

First, just focus on integrating the expression before evaluating it. When integrating, raise the exponent by one and then put that result on the denominator. Therefore, after integrating, you get: \(\displaystyle 3(\frac{x^4}{4})-2(\frac{x^3}{3})+x\). Then, plug in \(\displaystyle 2\), which yields \(\displaystyle \frac{26}{3}\) and then subtract the result of when you plug in \(\displaystyle 0\).

\(\displaystyle \frac{26}{3}-0=\frac{26}{3}\).

Example Question #151 : Writing Equations

\(\displaystyle \int (x+1)(x^3-2)dx\)

Possible Answers:

\(\displaystyle \frac{x^5}{5}+\frac{x^4}{4}-x^2+C\)

\(\displaystyle \frac{x^5}{5}+\frac{x^4}{4}-\frac{x^2}{2}+C\)

\(\displaystyle \frac{x^5}{5}+\frac{x^4}{4}-x^2\)

\(\displaystyle -\frac{x^5}{5}-\frac{x^4}{4}-x^2+C\)

Correct answer:

\(\displaystyle \frac{x^5}{5}+\frac{x^4}{4}-x^2+C\)

Explanation:

Before integrating, I would multiply the binomials (using FOIL) together to get a polynomial: \(\displaystyle \int x^4+x^3-2x-2dx\).  Now, integrate. Remember, when integrating, raise the exponent by 1 and then put that result on the denominator as well. You then get: \(\displaystyle \frac{x^5}{5}+\frac{x^4}{4}-\frac{2x^2}{2}\). Simplify and add a "C" at the end because it is an indefinite integral: \(\displaystyle \frac{x^5}{5}+\frac{x^4}{4}-x^2+C\).

Example Question #152 : Integral Expressions

\(\displaystyle \int \frac{4}{x}dx\)

Possible Answers:

\(\displaystyle ln\left | x \right |+C\)

\(\displaystyle 4ln\left | x \right |+C\)

\(\displaystyle 4ln\left | x \right |\)

\(\displaystyle 4ln\left |4 x \right |+C\)

Correct answer:

\(\displaystyle 4ln\left | x \right |+C\)

Explanation:

To integrate this expression, first take the 4 outside of the integral expression: \(\displaystyle 4\int \frac{1}{x}dx\). Recall that when you are integrating \(\displaystyle \frac{1}{x}\), the result is \(\displaystyle ln\left | x \right |\). Then, multiply by 4 and tack on a "C" because it is an indefinite integral: \(\displaystyle 4ln\left | x \right |+C\).

Example Question #151 : Integral Expressions

\(\displaystyle \int 2x^2-4x+1dx\)

Possible Answers:

\(\displaystyle \frac{2}{3}x^3-2x^2+2x+C\)

\(\displaystyle \frac{2}{3}x^3-2x^2+x+C\)

\(\displaystyle \frac{2}{3}x^3+2x^2+x+C\)

\(\displaystyle \frac{2}{3}x^3-2x^2+x\)

Correct answer:

\(\displaystyle \frac{2}{3}x^3-2x^2+x+C\)

Explanation:

To integrate this expression, raise the exponent by 1 and then put that result on the denominator (while multiplying that expression by any coefficients). Therefore, when you integrate, you get: \(\displaystyle \frac{2}{3}x^3-\frac{4x^2}{2}+x\). Then, simplify and add "C" because it is an indefinite integral so that your answer is: \(\displaystyle \frac{2}{3}x^3-2x^2+x+C\).

Example Question #154 : Integral Expressions

Evaluate the following Indefinte Integral:

\(\displaystyle \int(Sec^2(x))-3Sin(x))dx\)

Possible Answers:

\(\displaystyle Tan(x)-3Cos(x)+C\)

\(\displaystyle -Tan(x)+3Cos(x)+C\)

\(\displaystyle Tan(x)+3Cos(x)+C\)

\(\displaystyle Tan^2(x)+3Cos(x)+C\)

\(\displaystyle Tan(x)+Cos(x)+C\)

Correct answer:

\(\displaystyle Tan(x)+3Cos(x)+C\)

Explanation:

For this problem one must remember the following rule for Integrals: \(\displaystyle \int(f(x)+g(x)))dx=\int(f(x))dx+\int (g(x))dx\)

Once applying this rule we end with the following two integrals:

\(\displaystyle \int(Sec^2(x))-3Sin(x))dx=\int (Sec^2(x))dx+\int (3Sin(x))dx\)

 

Now apply our basic rules for Trigonometric Integrals we find the following:

\(\displaystyle Tan(x)+3Cos(x)+C\)

Note: a "+C" is nessecary as this integral is indefinite, that is, we are not plugging in any limits of integration.

Example Question #155 : Integral Expressions

Evaluate the following integral:

\(\displaystyle \int 2x^4e^{x^5}dx\)

Possible Answers:

\(\displaystyle -\frac{2}{5}e^{x^5}+C\)

\(\displaystyle \frac{2}{5}e^{x^5}\)

\(\displaystyle \frac{2}{5}e^{x^5}+C\)

\(\displaystyle 2e^{x^5}+C\)

Correct answer:

\(\displaystyle \frac{2}{5}e^{x^5}+C\)

Explanation:

To evaluate the integral, we must make the following substitution:

\(\displaystyle u=x^5, du=5x^4dx\)

Next, we rewrite the integral and integrate:

\(\displaystyle \frac{2}{5}\int e^u du=\frac{2}{5}e^u+C\)

The integral was performed using the following rule:

\(\displaystyle \int e^udu =e^u+C\)

Finally, replace u with our original term:

\(\displaystyle \frac{2}{5}e^{x^5}+C\)

Example Question #156 : Integral Expressions

Evaluate the following integral:

\(\displaystyle \int (\sec^2(x)+x^4+3x)dx\)

Possible Answers:

\(\displaystyle \tan(x)+x^5+\frac{3}{2}x^2+C\)

\(\displaystyle \tan(x)+\frac{x^5}{5}+{3}x^2+C\)

\(\displaystyle -\tan(x)+\frac{x^5}{5}+\frac{3}{2}x^2+C\)

\(\displaystyle \tan(x)+\frac{x^5}{5}+\frac{3}{2}x^2+C\)

Correct answer:

\(\displaystyle \tan(x)+\frac{x^5}{5}+\frac{3}{2}x^2+C\)

Explanation:

The integral is equal to

\(\displaystyle \tan(x)+\frac{x^5}{5}+\frac{3}{2}x^2+C\)

and was found using the following rules:

\(\displaystyle \int \sec^2(x)dx=\tan(x)+C\)\(\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C\)

Example Question #157 : Integral Expressions

Find the area under the curve \(\displaystyle y=5x^{4}+12x^{3}+ 9x^{2}+5\) between \(\displaystyle x=1\) and \(\displaystyle x=4\).

Possible Answers:

\(\displaystyle 12\)

\(\displaystyle 2004\)

\(\displaystyle 1992\)

\(\displaystyle -1992\)

\(\displaystyle -2004\)

Correct answer:

\(\displaystyle 1992\)

Explanation:

In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

\(\displaystyle \int_{a}^{b}f(x) dx\)

We plug in the equation of the curve in for \(\displaystyle f(x)\), the smaller of our x-values for \(\displaystyle a\), and the larger of our x-values for \(\displaystyle b\). For this problem, our set-up looks like this:

\(\displaystyle \int_{1}^{4}(5x^{4}+12x^{3}+ 9x^{2}+5 )dx\)

Next, we integrate our expression, ignoring our \(\displaystyle a\) and \(\displaystyle b\) values for now. We get:

\(\displaystyle F(x)= x^{5}+3x^{4}+3x^{3}+5x\)

Now, to find the area under the part of the curve we're looking for, we plug our \(\displaystyle a\) and \(\displaystyle b\) values into our integrated expression and find the difference, using the following skeleton:

\(\displaystyle area = F(b)-F(a)\)

In this problem, this looks like:

\(\displaystyle area = F(4)-F(1)\)

Which plugged into our integrated expression is:

\(\displaystyle area = [(4)^{5}+3(4)^{4}+3(4)^{3}+5(4)] - [(1)^{5}+3(1)^{4}+3(1)^{3}+5(1)]\)

\(\displaystyle area = [1024+768+192+20)] - [1+3+3+5]\)

\(\displaystyle area = [2004] - [12]\)

With our final answer being:

\(\displaystyle area = 1992\)

Example Question #158 : Integral Expressions

Find the area under the curve \(\displaystyle y=4x^{5}+15x^{2}+ 17x\) between \(\displaystyle x=0\) and \(\displaystyle x=2\).

Possible Answers:

\(\displaystyle -\frac{571}{6}\)

\(\displaystyle \frac{286}{3}\)

\(\displaystyle -222\)

\(\displaystyle \frac{761}{6}\)

\(\displaystyle 222\)

Correct answer:

\(\displaystyle \frac{286}{3}\)

Explanation:

In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

\(\displaystyle \int_{a}^{b}f(x) dx\)

We plug in the equation of the curve in for \(\displaystyle f(x)\), the smaller of our x-values for \(\displaystyle a\), and the larger of our x-values for \(\displaystyle b\). For this problem, our set-up looks like this:

\(\displaystyle \int_{0}^{2}(4x^{5}+15x^{2}+ 17x)dx\)

Next, we integrate our expression, ignoring our \(\displaystyle a\) and \(\displaystyle b\) values for now. We get:

\(\displaystyle F(x)= \frac{4}{6}x^{6}+5x^{3}+\frac{17}{2}x^{2}\)

Now, to find the area under the part of the curve we're looking for, we plug our \(\displaystyle a\) and \(\displaystyle b\) values into our integrated expression and find the difference, using the following skeleton:

\(\displaystyle area = F(b)-F(a)\)

In this problem, this looks like:

\(\displaystyle area = F(2)-F(0)\)

Which plugged into our integrated expression is:

\(\displaystyle area = [\frac{4}{6}(2)^{6}+5(2)^{3}+\frac{17}{2}+(2)^{2}] - [\frac{4}{6}(0)^{6}+5(0)^{3}+\frac{17}{2}(0)^{2}]\)

\(\displaystyle area = [\frac{4}{6}(64)+5(8)+\frac{17}{2}+(4)] - [\frac{4}{6}(0)+5(0)+\frac{17}{2}(0)]\)

\(\displaystyle area = [\frac{256}{6}+40+\frac{17}{2}+4] - [0+0+0]\)

\(\displaystyle area = [\frac{256}{6}+\frac{240}{6}+\frac{51}{6}+\frac{24}{6}] - [0]\)

\(\displaystyle area = [\frac{571}{6}] - [0]\)

With our final answer being:

\(\displaystyle area = \frac{571}{6}=\frac{286}{3}\)

Example Question #159 : Integral Expressions

Find the area under the curve \(\displaystyle y=15x\) between \(\displaystyle x=3\) and \(\displaystyle x=5\).

Possible Answers:

\(\displaystyle 30\)

\(\displaystyle \frac{375}{2}\)

\(\displaystyle 120\)

\(\displaystyle -30\)

\(\displaystyle -120\)

Correct answer:

\(\displaystyle 120\)

Explanation:

In order to find the area underneath a curve for a specific range of values, we need to set up a definite integral expression. To do this, we plug our specific information into the following skeleton:

\(\displaystyle \int_{a}^{b}f(x) dx\)

We plug in the equation of the curve in for \(\displaystyle f(x)\), the smaller of our x-values for \(\displaystyle a\), and the larger of our x-values for \(\displaystyle b\). For this problem, our set-up looks like this:

\(\displaystyle \int_{3}^{5}(15x)dx\)

Next, we integrate our expression, ignoring our \(\displaystyle a\) and \(\displaystyle b\) values for now. We get:

\(\displaystyle F(x)= \frac{15}{2}x^{2}\)

Now, to find the area under the part of the curve we're looking for, we plug our \(\displaystyle a\) and \(\displaystyle b\) values into our integrated expression and find the difference, using the following skeleton:

\(\displaystyle area = F(b)-F(a)\)

In this problem, this looks like:

\(\displaystyle area = F(5)-F(3)\)

Which plugged into our integrated expression is:

\(\displaystyle area = [\frac{15}{2}(5)^{2}] - [\frac{15}{2}(3)^{2}]\)

\(\displaystyle area = [\frac{15}{2}(25)] - [\frac{15}{2}(9)]\)

\(\displaystyle area = [\frac{375}{2}] - [\frac{135}{2}]\)

\(\displaystyle area = \frac{240}{2}\)

With our final answer being:

\(\displaystyle area = 120\)

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