Topics
>
Precalculus>
Trigonometric Functions>
Trig Applications of Integration>
Trig Applications of IntegrationTrig Applications of Integration
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 6 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
- 6
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- 5
- 6
- Answered
- Review
-
Question 1 of 6
1. Question
Find the area of the shaded regions formed by the curve `y=-cosx` that intercepts the x-axis at `x=pi/2`
Hint
Help VideoCorrect
Great Work!
Incorrect
Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`First, identify the limits of the shaded regionsFirst region`\text(Upper Limit)` `=` `pi/2` `\text(Lower Limit)` `=` `pi/4` $$\int_{\frac{\pi}{2}}^{\pi}\;-\text{cos}\;x\;dx$$ Second region`\text(Upper Limit)` `=` `pi` `\text(Lower Limit)` `=` `pi/2` $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\;-\text{cos}\;x\;dx$$ Next, integrate the trigonometric functionsFirst term$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\;-\text{cos}\;x\;dx$$ `=` $$-\bigg[\text{sin}\;x\bigg]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$ Integrate `\text(cos) x` Second term$$\int_{\frac{\pi}{2}}^{\pi}\;-\text{cos}\;x\;dx$$ `=` $$-\bigg[\text{sin}\;x\bigg]_{\frac{\pi}{2}}^{\pi}$$ Integrate `\text(cos)x` Next, get the difference of the upper and lower limits substituted to the integral as `x` for each term.First term$$\bigg[-\text{sin}\;x\bigg]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$ `=` `-\text(sin)(pi/2)-[-\text(sin)(pi/4)]` Substitute the limits `=` `-1+1/(sqrt2)``times(-1)` Multiply the values by `-1` to make it positive `=` `1-1/(sqrt2)` Second term$$\bigg[-\text{sin}\;x\bigg]_{\frac{\pi}{2}}^{\pi}$$ `=` `-\text(sin)(pi)-[-\text(sin)(pi/2)]` Substitute the limits `=` `0+1` Evaluate `=` `1` Finally, add the values of the first and second term to get the area of the shaded regions`\text(Area)` `1-1/(sqrt2)``+``1` `=` `2-1/(sqrt2)` `2-1/(sqrt2)` -
Question 2 of 6
2. Question
Find the area of the shaded regions formed by the curves `y=sin2x` and `y=sinx` that intercepts at `x=pi/3`
Write fractions in the format “a/b”- `A=` (1/4)` \text(units)^2`
Hint
Help VideoCorrect
Correct!
Incorrect
Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`First, identify the limits of the shaded region`\text(Upper Limit)` `=` `pi/3` `\text(Lower Limit)` `=` `0` $$\int_{0}^{\frac{\pi}{3}}\;\text{sin}\;2x-\text{sin}\;x\;dx$$ Next, integrate the trigonometric function$$\int_{0}^{\frac{\pi}{3}}\;\text{sin}\;2x-\text{sin}\;x\;dx$$ `=` $$\bigg[-\frac{1}{2}\text{cos}\;2x-(-\text{cos}\;x)\bigg]_{0}^{\frac{\pi}{3}}$$ Integrate `\text(sin) 2x-\text(sin) x` `=` $$\bigg[-\frac{1}{2}\text{cos}\;2x+\text{cos}\;x\bigg]_{0}^{\frac{\pi}{3}}$$ Simplify Finally, get the difference of the upper and lower limits substituted to the integral as `x` for each term.$$\bigg[-\frac{1}{2}\text{cos}\;2x+\text{cos}\;x\bigg]_{0}^{\frac{\pi}{3}}$$ `=` `[-1/2\text(cos) 2(pi/3)+\text(cos) pi/3]-[-1/2\text(cos) 2(0)+\text(cos) 0]` Substitute the limits `=` `[(-1/2*-1/2)+1/2]-[-1/2+1]` Substitute known trigonometric values `=` `1/4+1/2+1/2-1` Evaluate `=` `1/4` `A=1/4 \text(units)^2` -
Question 3 of 6
3. Question
Find the area of the shaded regions formed by the curve `y=cosx` that intercepts the straight line `y=1/2` and the x-axis at `x=pi/2`
Hint
Help VideoCorrect
Keep Going!
Incorrect
Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`First, find the value of the x-intercept`y` `=` `\text(cos) x` `y` `=` `1/2` `\text(cos) x` `=` `1/2` `x` `=` `pi/3` Compute using calculator Next, solve the red-shaded region by using the formula for area of a rectangle`\text(length)` `=` `x` `=` `pi/3` `\text(height)` `=` `y` `=` `1/2` `\text(Area)` `=` `\text(length)xx\text(height)` `=` `pi/3xx1/2` Substitute known values `=` `pi/6` Next, identify the limits of the yellow-shaded region`\text(Upper Limit)` `=` `pi/2` `\text(Lower Limit)` `=` `pi/` $$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\;\text{cos}\;x\;dx$$ Then, integrate the trigonometric function$$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\;\text{cos}\;x\;dx$$ `=` $$\bigg[\text{sin}\;x\bigg]_{\frac{\pi}{3}}^{\frac{\pi}{2}}$$ Integrate `\text(cos) x` Next, get the difference of the upper and lower limits substituted to the integral as `x` for each term.$$-\bigg[\text{sin}\;x\bigg]_{\frac{\pi}{2}}^{\pi}$$ `=` `\text(sin)(pi/2)-[\text(sin)(pi/3)]` Substitute the limits `=` `1-(sqrt3)/2` Substitute known trigonometric values Finally, add the areas of the rectangle and the yellow-shaded region term to get the total area`\text(Area)` `pi/6``+``1-(sqrt3)/2` `A=pi/6+1-(sqrt3)/2 \text(units)^2` -
Question 4 of 6
4. Question
Find the area of the shaded regions formed by the curve `y=cosx` that intercepts the x-axis at `x=pi/2` and the straight line `y=2x+1` that intercepts the x-axis at `x=-1/2`
Hint
Help VideoCorrect
Fantastic!
Incorrect
Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`First, identify the limits of the shaded regionsYellow region`\text(Upper Limit)` `=` `0` `\text(Lower Limit)` `=` `-1/2` $$\int_{-\frac{1}{2}}^{0}\;(2x+1)\;dx$$ Green region`\text(Upper Limit)` `=` `pi/2` `\text(Lower Limit)` `=` `0 $$\int_{0}^{\frac{\pi}{2}}\;\text{cos}\;x\;dx$$ Next, integrate the trigonometric functionsYellow region$$\int_{-\frac{1}{2}}^{0}\;(2x+1)\;dx$$ `=` $$-\bigg[x^2+x\bigg]_{-\frac{1}{2}}^{0}$$ Integrate `2x+1` Green region$$\int_{0}^{\frac{\pi}{2}}\;\text{cos}\;x\;dx$$ `=` $$\bigg[\text{sin}\;x\bigg]_{0}^{\frac{\pi}{2}}$$ Integrate `\text(cos)x` Next, get the difference of the upper and lower limits substituted to the integral as `x` for each term.Yellow region$$-\bigg[x^2+x\bigg]_{-\frac{1}{2}}^{0}$$ `=` `[(0)^2+0]-[(-1/2)^2+1/2]` Substitute the limits `=` `-(1/4-1/2)` Evaluate `=` `-(-1/4)` `=` `1/4` Green region$$\bigg[\text{sin}\;x\bigg]_{0}^{\frac{\pi}{2}}$$ `=` `[\text(sin) (pi)/2]-[\text(sin)(0)]` Substitute the limits `=` `1-0` `\text(sin) pi/2=1` `=` `1` Finally, add the values of the shaded regions to get the total area`\text(Area)` `1/4``+``1` `=` `1 1/4` `A=1 1/4 \text(units)^2` -
Question 5 of 6
5. Question
A region formed by the curve `y=2sec2x` and x-axis is rotated about the x-axis from `x=0` to `x=pi/6`. Find the volume of this solid of revolution.
Hint
Help VideoCorrect
Excellent!
Incorrect
Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`Volume of a Solid of Revolution
$$V=\int_{\color{#D800AD}{a}}^{\color{#9a00c7}{b}}\;\pi\color{#004ec4}{y}^2\;dx$$First, identify the upper and lower limits`\text(Upper Limit)` `=` `pi/6` `\text(Lower Limit)` `=` `0` Next, substitute the known values to the volume formula`V` `=` $$\int_{\color{#D800AD}{a}}^{\color{#9a00c7}{b}}\;\pi\color{#004ec4}{y}^2\;dx$$ `=` $$\int_{\color{#D800AD}{0}}^{\color{#9a00c7}{\frac{\pi}{6}}}\;\pi\cdot(\color{#004ec4}{2\text{sec}})^2\;2x\;dx$$ Substitute known values `=` $$\int_{0}^{\frac{\pi}{6}}\;\pi\cdot 4\text{sec}^2\;2x\;dx$$ Evaluate `=` $$4\pi\int_{0}^{\frac{\pi}{6}}\;\text{sec}^2\;2x\;dx$$ `=` $$4\pi\bigg[\frac{1}{2}\;\text{tan}\;2x\bigg]_{0}^{\frac{\pi}{6}}$$ Integrate `=` $$2\pi\bigg[\text{tan}\;2x\bigg]_{0}^{\frac{\pi}{6}}$$ Simplify `=` $$2\pi\bigg[\text{tan}\;2\left(\color{#9a00c7}{\frac{\pi}{6}}\right)-\text{tan}\;2(\color{#D800AD}{0})\bigg]$$ Substitute the limits `=` $$2\pi\bigg[\text{tan}\left(\frac{\pi}{3}\right)-0\bigg]$$ `=` $$2\pi\cdot\sqrt3$$ `\text(tan) pi/3=sqrt3` `=` $$2\pi\sqrt3$$ `V=2pisqrt3 \text(units)^3` -
Question 6 of 6
6. Question
A region formed by the curve `y=sqrt(sinx)` and x-axis is rotated about the x-axis from `x=0` to `x=pi`. Find the volume of this solid of revolution.
Hint
Help VideoCorrect
Nice Job!
Incorrect
Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`Volume of a Solid of Revolution
$$V=\int_{\color{#D800AD}{a}}^{\color{#9a00c7}{b}}\;\pi\color{#004ec4}{y}^2\;dx$$First, remove the surd from the function`y` `=` `sqrt(sinx)` `y^2` `=` `(sqrt(sinx))^2` Raise both sides by `2` `y^2` `=` `sinx` Next, identify the upper and lower limits`\text(Upper Limit)` `=` `pi` `\text(Lower Limit)` `=` `0` Next, substitute the known values to the volume formula`V` `=` $$\int_{\color{#D800AD}{a}}^{\color{#9a00c7}{b}}\;\pi\color{#004ec4}{y}^2\;dx$$ `=` $$\int_{\color{#D800AD}{0}}^{\color{#9a00c7}{\pi}}\;\pi\cdot(\color{#004ec4}{\text{sin}})\;x\;dx$$ Substitute known values `=` $$\pi\bigg[-\text{cos}\;x\bigg]_{0}^{\pi}$$ Integrate `=` $$\pi[- \text{cos} (\color{#9a00c7}{\pi}) ]-[-\text{cos}(\color{#D800AD}{0})]$$ Substitute the limits `=` $$\pi[-(-1)-(-1)]$$ Substitute known trigonometric values `=` $$\pi(1+1)$$ `=` $$2\pi$$ `V=2pi \text(units)^3`
Quizzes
- Derivative of Trigonometric Functions 1
- Derivative of Trigonometric Functions 2
- Derivative of Trigonometric Functions 3
- Trig Applications of Differentiation
- Integral of Trigonometric Functions 1
- Integral of Trigonometric Functions 1
- Trig Applications of Integration
- Graph Trigonometric Functions 1
- Graph Trigonometric Functions 2
- Graph Trigonometric Functions 3
- Graph Trigonometric Functions 4