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Flashcards: Find the Area Between Curves

Determine the area of the region bounded by y=x^2+1 $\displaystyle y=x^2+1$ and $\displaystyle y=sin(x)$, with $0\leq x\leq 1$ $\displaystyle 0\leq x\leq 1$. Round to the nearest hundredth.

-0.21 $\displaystyle -0.21$

1.17 $\displaystyle 1.17$

0.87 $\displaystyle 0.87$

2.87 $\displaystyle 2.87$

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Flashcards: Find the Area Between Curves

Determine the area of the region bounded by y=x^2+1 $\displaystyle y=x^2+1$ and $\displaystyle y=sin(x)$, with $0\leq x\leq 1$ $\displaystyle 0\leq x\leq 1$. Round to the nearest hundredth.

-0.21 $\displaystyle -0.21$

1.17 $\displaystyle 1.17$

0.87 $\displaystyle 0.87$

2.87 $\displaystyle 2.87$

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Flashcards: Find the Area Between Curves

Determine the area of the region bounded by y=x^2+1 $\displaystyle y=x^2+1$ and $\displaystyle y=sin(x)$, with $0\leq x\leq 1$ $\displaystyle 0\leq x\leq 1$. Round to the nearest hundredth.

Your answer:

-0.21 $\displaystyle -0.21$

Your answer:

1.17 $\displaystyle 1.17$

Your answer:

2.87 $\displaystyle 2.87$

Correct answer:

0.87 $\displaystyle 0.87$

Explanation:

In this particular problem, the bounds for our integral are provided; specifically, we know that the integral setup should look something like $\int _0^1(upper function)-(lower function) dx$ $\displaystyle \int _0^1(upper function)-(lower function) dx$. This can be determined by the given domain $0\leq x\leq 1$ $\displaystyle 0\leq x\leq 1$ and the fact that we are dealing with an area problem involving two functions.

The next step is to continue creating the integral that will be used to evaluate the area. By graphing the functions, it can be determined that the function y=x^2+1 $\displaystyle y=x^2+1$ takes on larger values of $\displaystyle y$ than the other function $\displaystyle y=sin(x)$, and therefore should be deemed the “upper function.”

Without graphing, this can also be determined because $\displaystyle y=sin(x)$ will take on values between $\displaystyle 0$ and $\displaystyle 1$, while x^2+11 $\displaystyle x^2+11$.

The following integral is then created:

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

Finally, the definite integral is evaluated:

$\int_{0}^{0}(x^2+1)-(sin(x)) dx$ $\displaystyle \int_{0}^{0}(x^2+1)-(sin(x)) dx$ $=[\frac{1}{3}x^3+x+cos(x)] |_0^1=(\frac{1}{3}+1+cos(1))-(0+0+1)=\frac{1}{3}+cos(1)=0.87$ $\displaystyle =[\frac{1}{3}x^3+x+cos(x)] |_0^1=(\frac{1}{3}+1+cos(1))-(0+0+1)=\frac{1}{3}+cos(1)=0.87$

Flashcards: Find the Area Between Curves

Determine the area of the region bounded by y=5x^3 $\displaystyle y=5x^3$ and y=3x^2 $\displaystyle y=3x^2$. Round to the nearest hundredth.

0.05 $\displaystyle 0.05$

0.38 $\displaystyle 0.38$

0.25 $\displaystyle 0.25$

$\displaystyle 0$

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Flashcards: Find the Area Between Curves

Determine the area of the region bounded by y=5x^3 $\displaystyle y=5x^3$ and y=3x^2 $\displaystyle y=3x^2$. Round to the nearest hundredth.

Your answer:

0.38 $\displaystyle 0.38$

Your answer:

0.25 $\displaystyle 0.25$

Your answer:

$\displaystyle 0$

Correct answer:

0.05 $\displaystyle 0.05$

Explanation:

First, the intersection points of the two functions must be found in order to determine the bounds of the area integral we wish to set up.

y=5x^3 $\displaystyle y=5x^3$ and y=3x^2 $\displaystyle y=3x^2$

$\displaystyle 5x^3=3x^2$

$\displaystyle x={0,0.6}$

Next, the area integral must be set up correctly. To do this, the two functions must be either categorized as an “upper function” or a “lower function.” This is important as it determines which function comes first in our general formula for the area between two curves, $\int_{a}^{b} (upper function)-(lower function) dx$ $\displaystyle \int_{a}^{b} (upper function)-(lower function) dx$.

After examination either numerically or graphically, y=3x^2 $\displaystyle y=3x^2$ is determined to be above the other function, y=5x^3 $\displaystyle y=5x^3$. Now, the integral may be set up:

$\int_{0}^{0.6} (3x^2)-(5x^3) dx$ $\displaystyle \int_{0}^{0.6} (3x^2)-(5x^3) dx$

Finally, the integral must be evaluated:

$Area = \int_{0}^{0.6}(3x^2)-(5x^3) dx = [x^3-\frac{5}{4}x^4] |_0^{0.6} = (0.6)^3-\frac{5}{4}(0.6)^4 = 0.05$ $\displaystyle Area = \int_{0}^{0.6}(3x^2)-(5x^3) dx = [x^3-\frac{5}{4}x^4] |_0^{0.6} = (0.6)^3-\frac{5}{4}(0.6)^4 = 0.05$

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Flashcards: Find the Area Between Curves

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Flashcards: Find the Area Between Curves

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