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Calculus 1 : Equations

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Example Questions

Example Question #1 : Equations

What is the area under the curve 2^x for x < 0 ?

Possible Answers:

$\frac{1}{\ln 2}$

e^2

$\ln 2$

Correct answer:

$\frac{1}{\ln 2}$

Explanation:

By normal exponent rules $2^x = e^{x \ln 2}$ , and so we set up the definite integral

$\int_{-\infty}^{0} e^{x \ln 2} ~dx$ ,

which integrates to:

$\left[ \frac{e^{x \ln 2}}{\ln 2}\right]_{-\infty}^0 = \frac{2^0 - 2^{-\infty}}{\ln 2} = \frac{1 - 0}{\ln 2}$

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Example Question #2 : Equations

The Gaussian integral formula states that

$\int_{-\infty}^{\infty} e^{-x^2} ~ dx = \sqrt{\pi}$ .

What is the integral of

$\int_{-\infty}^{\infty} x^2 ~ e^{-x^2} ~ dx$ ?

Possible Answers:

$\frac{\pi}{4}$

$\frac{\sqrt{\pi}}{2}$

$\sqrt{\pi}$

$\sqrt{2\pi}$

$\sqrt{\frac{\pi}{2}}$

Correct answer:

$\frac{\sqrt{\pi}}{2}$

Explanation:

Integrating by parts with

u = x , du = dx

$v = - \frac{1}{2} e^{-x^2}$ , $dv = x ~e^{-x^2}~dx$ ,

we get:

$\int_{-\infty}^{\infty} x^2 e^{-x^2} dx = \int_A^B u ~ dv = \left[ u v\right ]_A^B - \int_A^B v ~ du$

$= \left[ \frac{-x}{2} e^{-x^2} \right ]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{-1}{2} e^{-x^2}~ dx = [0 - 0] - \frac{-1}{2} \sqrt{\pi} = \frac{\sqrt{\pi}}{2}$

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Example Question #3 : Equations

Determine f(x) given that f'(x)=15x^4+14x^3-9x^2+5x-16 .

Possible Answers:

$3x^5+\frac{7}{2}x^4-3x^3+\frac{5}{2}x^2-16x+C$

$-3x^5-\frac{7}{2}x^4+3x^3-\frac{5}{2}x^2+16x$

$\frac{15}{4}x^5+\frac{14}{3}x^4-18x^3+5x^2-16x+C$

$-\frac{15}{4}x^5-\frac{14}{3}x^4+18x^3-5x^2+16x$

5x^3+42x^4-18x^3+5x^2-32x+C

Correct answer:

$3x^5+\frac{7}{2}x^4-3x^3+\frac{5}{2}x^2-16x+C$

Explanation:

This problem requires you to evaluate an indefinite integral of the given function f’(x). Integrating each term of the function with respect to x, we simply divide each coefficient by (n+1), where n is the value of the exponent on x for that particular term, and then add 1 to the value of the exponent on each x. This gives us:

$f(x)=\int f'(x)dx=\int (15x^4+14x^3-9x^2+5x-16)dx$

$f(x)=3x^5+\frac{7}{2}x^4-3x^3+\frac{5}{2}x^2-16x+C$

A correct answer must include a constant C, as the original function may have had a constant that is not reflected in the equation for its derivative.

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Example Question #4 : Equations

What is the value of the following definite integral?

$\int_{0}^{\pi /2}\bigg(-5sin(x)-3cos(x)+\frac{4}{\pi }\bigg)dx$

Possible Answers:

$-5-3cos\bigg(\frac{\pi }{2}\bigg)$

$5cos\bigg(\frac{\pi }{2}\bigg)+3$

Correct answer:

Explanation:

We start by integrating the function in parentheses with respect to x, and we then subtract the evaluation of the lower limit from the evaluation of the upper limit:

$\int_{0}^{\pi /2}\bigg(-5sin(x)-3cos(x)+\frac{4}{\pi })dx=(5cos(x)-3sin(x)+\frac{4}{\pi }x\bigg)_{0}^{\pi /2}$

$=(5cos(\frac{\pi }{2})-3sin(\frac{\pi }{2})+\frac{4}{\pi }(\frac{\pi }{2}))-(5cos(0)-3sin(0)+\frac{4}{\pi}(0))$

=(-3+2)-(5)=-6

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Example Question #5 : Equations

Determine the amount of work required to push a box from x=2 meters to x=5 meters, given the function for force below:

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

Possible Answers:

$39.8 \text{ J}$

$19.8 \text{ J}$

$14.2 \text{ J}$

$43.2 \text{ J}$

$78.9\text{ J}$

Correct answer:

$39.8 \text{ J}$

Explanation:

This problem intends to demonstrate one possible application of calculus to the field of physics. An equation for work as given in physics is as follows:

$W=\int F(x)dx$

Using this equation, we simply set up a definite integral with our bounds given by the interval of interest in the problem:

$W=\int_{2}^{5}\bigg(x^2-sin(2x)+\frac{1}{x}\bigg)dx$

$W=\bigg(\frac{1}{3}x^3+\frac{1}{2}cos(2x)+ln(x)\bigg)_{2}^{5}$

$W=\bigg(\frac{1}{3}(5)^3+\frac{1}{2}cos(10)+ln(5)\bigg)-\bigg(\frac{1}{3}(2)^3+\frac{1}{2}cos(4)+ln(2)\bigg)$

W=39.8 Joules

After evaluating our integral, we can see that the work required to push the box from x=2 meters to x=5 meters is 39.8 Joules.

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Example Question #6 : Equations

Evaluate the definite integral within the interval $0\leq x \leq 1$ .

$\int(5x^{4}+x^{2}-x)dx$

Possible Answers:

$\frac{5}{6}$

$-\frac{5}{6}$

Correct answer:

$\frac{5}{6}$

Explanation:

In order to solve this problem we must know that

$\int_{a}^{b}f(x)=F(b)-F(a)$

In this case we have:

$\int_{0}^{1} (5x^{4}+x^{2}-x)dx$

Our first step is to integrate:

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

We then arrive to our solution by plugging in our values a=0, b=1

$\left [ (1)^{5}+\frac{1}3{(1)^{3}-\frac{1}{2}(1)^{2}} \right ]-\left [ (0)^{5}+\frac{1}{3}(0)^{3}-\frac{1}{2}(0)^{2} \right ]$

$\left ( 1+\frac{1}{3}-\frac{1}{2} \right )-0=\frac{5}{6}$

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Example Question #7 : Equations

Write out the expression of the area under the following curve from x=1 to x=4 .

y=6x^2+4x+5

Possible Answers:

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

$\int_{1}^{4}(6x^2+4x+5)$

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

$\int_{1}^{4}(6x^2+4x+5)dx$

Correct answer:

$\int_{1}^{4}(6x^2+4x+5)dx$

Explanation:

Simply use an integral expression with the given x values as your bounds.

Don't forget to add the dx!

$\int_{1}^{4}(6x^2+4x+5)dx$

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Example Question #8 : Equations

Evaluate the definite integral within the interval $1\leq x \leq 2$ .

$\int (2x^{3}+\frac{1}{2}x^{2}+7)dx$

Possible Answers:

$\frac{47}{3}$

$-\frac{47}{3}$

$\frac{31}{2}$

$-\frac{31}{2}$

Correct answer:

$\frac{47}{3}$

Explanation:

In order to solve this problem we must remember that:

$\int_{a}^{b}f(x)=F(b)-F(a)$

In this case we have:

$\int_{1}^{2}(2x^{3}+\frac{1}{2}x^{2}+7) dx$

Our first step is to integrate:

$\int (2x^3+\frac{1}{2}x^2+7)dx= \frac{1}{2}x^4+\frac{1}{6}x^3+7x$

We then arrive at our solution by plugging in our values a=1, b=2

$\left[\frac{1}{2}(2)^4+\frac{1}{6}(2)^3+7(2)\right]-\left[\frac{1}{2}(1)^4+\frac{1}{6}(1)^3+7(1)\right]$

$\left ( 8+\frac{4}{3} +14 \right)-\left(\frac{1}{2}+\frac{1}{6}+7 \right)=\frac{47}{3}$

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Example Question #9 : Equations

Evaluate the definite integral within the interval $4\leq x \leq 9$

$\int \left(\frac{1}{\sqrt{x}}+3x^2-4\right)dx$

Possible Answers:

647

$\frac{-1169}{6}$

$\frac{1169}{6}$

Correct answer:

647

Explanation:

In order to solve this problem we must remember that:

$\int_{a}^{b}f(x)=F(b)-F(a)$

In this case we have:

$\int_{4}^{9}\left(\frac{1}{\sqrt{x}}+3x^2-4\right)dx$

Our first step is to integrate:

$\int\left(\frac{1}{\sqrt{x}}+3x^2-4\right)dx=2\sqrt{x}+x^{3}-4x$

We then arrive to our solution by plugging in our values a=4, b=9

$\left [ 2\sqrt{9}+(9)^{3}-4(9) \right ]-\left [ 2\sqrt{4}+(4)^{3}-4(4) \right ]$

(6+729-36)-(4+64-16)

699-52=647

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Example Question #4 : Integral Expressions

Evaluate the indefinite integral.

$\int (5+4x)dx$

Possible Answers:

5x+2x^2+c

4x^2+5x+c

8x^2+5x+c

4+c

Correct answer:

5x+2x^2+c

Explanation:

We are being asked to integrate the function.

To do this we need to remember the power rule of integrals,

$\int (a+bx)dx=ax+\frac{bx^n+1}{n+1}+c$

Using this rule, we can evaluate the following integral:

$\int (5+4x)dx = 5x+2x^2+c$

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : Equations

Example Question #2 : Equations

Example Question #3 : Equations

Example Question #4 : Equations

Example Question #5 : Equations

Example Question #6 : Equations

Example Question #7 : Equations

Example Question #8 : Equations

Example Question #9 : Equations

Example Question #4 : Integral Expressions

All Calculus 1 Resources

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