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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Derivatives

Screen shot 2015 07 10 at 8.27.12 pm

Which of the following is true about the twice-differentiable function above?

Possible Answers:

f(2)<f'(2)<f''(2)

f(2)<f''(2)<f'(2)

f''(2)<f'(2)<f(2)

f'(2)<f''(2)<f(2)

f''(2)<f(2)<f'(2)

Correct answer:

f(2)<f''(2)<f'(2)

Explanation:

Since the function is increasing at x=2 , f'(2)>0 , and since f(2) is below the x-axis, f(2)<0 .

Furthermore, there exists an inflection point at x=2 , where the concavity of function changes.

Thus at x=2 , f''(2)=0 .

Therefore, the correct answer is f(2)<f''(2)<f'(2) .

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Example Question #5 : How To Graph Functions Of Curves

Which one of the following could be the integral of $cos2x+5\sqrt{x}$ ?

Possible Answers:

$\frac{\sin2x}{2}+\frac{10x^\frac{3}{2}}{3}+9$

$\sin2x+10x^\frac{3}{2}$

$\frac{\sin2x}{2}+\frac{10x^\frac{3}{2}}{3}+3x$

$-2{\sin2x}+\frac{1}{2\sqrt{x}}$

$\frac{\sin(x)}{2}+\frac{10x^\frac{3}{2}}{3}+9$

Correct answer:

$\frac{\sin2x}{2}+\frac{10x^\frac{3}{2}}{3}+9$

Explanation:

Since the functions are added together, we can take each one seperately and add the results together.

The integral of $\cos2x$ is $\frac{\sin2x}{2}$ since you must apply for the chain rule of the .

The integral of $5\sqrt{x}$ will be using the power rule

$\int x^n\rightarrow\frac{x^{n+1}}{n+1}$ ,

which means it will equal

$5\cdot \left(\frac{x^{\frac{3}{2}}}{(3/2)}\right)$ , which turns into $\frac{10x^\frac{3}{2}}{3}$ ,

so combining these gives

$\frac{\sin2x}{2}+\frac{10x^\frac{3}{2}}{3}+C$

as the integral, making $\frac{\sin2x}{2}+\frac{10x^\frac{3}{2}}{3}+9$ the only equation that satisfies this quality.

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Example Question #3 : How To Graph Functions Of Curves

Exponential Function

What is the graph of the folloiwng function:

y=x^3+5x^2+3x+9

Possible Answers:

None of the above

Correct answer:

Explanation:

Use the following values to plot the graph:

y=0 , x=0 , x=-3 , y=2

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Example Question #2 : How To Graph Functions Of Curves

Trigonometric Function

Graph the folloiwng function:

$9y-\cot x=0$

Possible Answers:

None of the above

Correct answer:

Explanation:

Plot the graph for the following values:

$x=0,x=\pm\frac{\pi}{2},y=\pm4$

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Example Question #1 : How To Graph Functions Of Area

Screen shot 2015 07 10 at 12

Graph of a piecewise-linear function , for [-2,4] , is shown above.

Find

$\int_{-2}^{4}f(x)dx$ .

Possible Answers:

-3.5

3.5

Correct answer:

Explanation:

Find the area under the graph from [-2,4] . To do this break the graph up into triangles, squares, and rectangles to calculate the individual areas over smaller intervals and then add them all together.

The areas are added to be:

$A=\left[\frac{1}{2}(1)(2)+\frac{1}{2}(1)(-1)+\frac{1}{2}(1)(-1)+(1)(1)+(2)(3)\right]=7$

Therefore,

$\int_{-2}^{4}f(x)dx=7$ .

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Example Question #2 : How To Graph Functions Of Area

Find the area bounded by the curve $y=\sin x \cos x$ and the -axis over the interval $\left[0,\frac{\pi}{2}\right]$ .

Possible Answers:

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{2}$

Explanation:

The curve is in quadrant one over the given interval, which gives us the bounds of integration. Evaluating this definite integral yields the area we are after.

In order to performe the antiderivative, let $u=\sin x$ . It follows that $du = \cos x dx$ . Therefore

$\int (\sin x \cos x)dx= \int u du=\frac{1}{2}u^2 + C = \frac{1}{2} \sin^2 x + C$

$\int_{0}^{\frac{\pi}{2}}(\sin x \cos x)dx= \left[ \frac {1}{2}\sin^2 x \right] ^\frac{\pi}{2}_0= \left( \frac{1}{2}\sin^2\frac{\pi}{2} \right) - \left( \frac{1}{2}\sin^2 0 \right)=\frac{1}{2}$

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Example Question #3 : How To Graph Functions Of Area

Find the area bounded by the curve y=e^x and the -axis over the interval [0,1] .

Possible Answers:

e+1

e-1

$\ln (e+1)$

Correct answer:

e-1

Explanation:

The curve is positive over the given interval, so the endpoints of the interval will mark the bounds of integration. This function is very easy to integrate because the derivative of e^x is itself!

$\int_{0}^{1} e^xdx=\left[ e^x\right]^1_0=e^1-e^0=e-1$

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Example Question #4 : How To Graph Functions Of Area

Find the area bounded by the curve $y=e^{-x}$ and the -axis over the interval [0,1] .

Possible Answers:

$\frac{1-e}{e}$

$\frac{e-1}{e}$

$\frac{1}{e}+1$

$\frac{e+1}{e}$

$\frac{1}{e}-1$

Correct answer:

$\frac{e-1}{e}$

Explanation:

This function is positve over the given interval, so the endpoints of the interval mark the bounds of integration. It is a straightforward integration that is solvable with u-substitution. Let u = -x so du = -dx . This means

$\int e^{-x}dx = - \int e^udu = -e^u + C = -e^{-x} + C$

$\int_{0}^{1} e^{-x}dx=\left[ -e^{-x}\right]^1_0=(-e^{-1})-(-e^0)=-\frac{1}{e}+1=\frac{e-1}{e}$

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Example Question #5 : How To Graph Functions Of Area

Find the area bounded by the curve y=2x-x^2 in the first quadrant.

Possible Answers:

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{3}{2}$

$\frac{2}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

The curve is in quadrant one over the interval [0,2] , which are the bounds of integration. To see this, note that the x-intercepts are 0 and 2 and the parabola opens downward.

The definite integral below is solved by taking the antiderivative of each term of the given polynomial function, evaluating this antiderivative at the bounds of integration, and subtracting the values.

For this particular integral use the rule, $\int x^n=\frac{x^{n+1}}{n+1}$ to solve.

$\int_{0}^{2}(2x-x^2)dx=(x^2-\frac{1}{3}x^3)|^2_0=(2^2-\frac{1}{3}\cdot2^3)-(0^2-\frac{1}{3}\cdot0^3)=\frac{4}{3}$

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Example Question #1 : Area

Find the area under the curve between the following bounderies of the following function.

h(x)=3x^2+7 in between the boundaries of and

Possible Answers:

205

125

160

100

Correct answer:

160

Explanation:

We can find the area under the curve by taking the anti-derivative of the function and using the two boundaries as x values. The anti-derivative of h(x)=3x^2+7 is x^3+7x . If we use the two boundaries, we end up with our answer, 5^3+7(5)-(0^3+7(0)) = 160 .

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : Derivatives

Example Question #5 : How To Graph Functions Of Curves

Example Question #3 : How To Graph Functions Of Curves

Example Question #2 : How To Graph Functions Of Curves

Example Question #1 : How To Graph Functions Of Area

Example Question #2 : How To Graph Functions Of Area

Example Question #3 : How To Graph Functions Of Area

Example Question #4 : How To Graph Functions Of Area

Example Question #5 : How To Graph Functions Of Area

Example Question #1 : Area

All Calculus 1 Resources

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