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Definite Integrals of Logarithmic FunctionsDefinite Integrals of Logarithmic Functions
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Question 1 of 6
1. Question
Find the definite integral$$\int_{0}^{2} \frac{x}{x^2+1}$$Hint
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Great Work!
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Integral of a Basic Fraction
$$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$$General Form Integral of a Fraction
$$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$$First, form a fraction to balance the equation.$$\frac{x}{\color{#CC0000}{x^2+1}}$$ `=` $$\frac{x}{\color{#CC0000}{2x}}$$ Differentiate `x^2+1` `=` `1/2` `x/x=1` Use `1/2` as a constant to balance the integral.`f(x)` `=` `x^2+1` `f'(x)` `=` `2x` $$\frac{1}{2}\int_{0}^{2} \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$$ `=` $$\frac{1}{2}\int_{0}^{2} \frac{\color{#9a00c7}{2x}}{\color{#D800AD}{x^2+1}}dx$$ Substitute known values `=` `[1/2 ln (x^2+1)]_0^2` Finally, get the difference of the upper and lower limits substituted to the integral as `x`.`[1/2 ln (x^2+1)]_0^2` `=` `1/2[ln ((2)^2+1)-ln((0)^2+1)]` Substitute the limits `=` `1/2[ln 5-ln1]` Evaluate `=` `1/2 ln 5` `1/2 ln 5` -
Question 2 of 6
2. Question
Find the definite integral$$\int_{1}^{7} \frac{x^2}{x^3+2}$$Round your answer to two decimal places- (1.58)
Hint
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Correct!
Incorrect
Integral of a Basic Fraction
$$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$$General Form Integral of a Fraction
$$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$$First, form a fraction to balance the equation.$$\frac{x^2}{\color{#CC0000}{x^3+2}}$$ `=` $$\frac{x^2}{\color{#CC0000}{3x^2}}$$ Differentiate `x^3+2` `=` `1/3` `x^2/x^2=1` Use `1/3` as a constant to balance the integral.`f(x)` `=` `x^3+2` `f'(x)` `=` `3x^2` $$\frac{1}{3}\int_{1}^{7} \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$$ `=` $$\frac{1}{3}\int_{1}^{7} \frac{\color{#9a00c7}{3x^2}}{\color{#D800AD}{x^3+2}}dx$$ Substitute known values `=` `[1/3 ln (x^3+2)]_1^7` Finally, get the difference of the upper and lower limits substituted to the integral as `x`.`[1/3 ln (x^3+2)]_1^7` `=` `1/3[ln ((7)^3+2)-ln((1)^3+2)]` Substitute the limits `=` `1/3[ln 347-ln 3]` Evaluate `=` `1/3[4.7507]` Compute using calculator `=` `1.58` Rounded to two decimal places `1.58` -
Question 3 of 6
3. Question
Find the definite integral$$\int_{1}^{3} \frac{1}{2x-1}$$Hint
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Keep Going!
Incorrect
Integral of a Basic Fraction
$$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$$General Form Integral of a Fraction
$$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$$First, form a fraction to balance the equation.$$\frac{1}{\color{#CC0000}{2x-1}}$$ `=` $$\frac{1}{\color{#CC0000}{2}}$$ Differentiate `2x-1` Use `1/2` as a constant to balance the integral.`f(x)` `=` `2x-1` `f'(x)` `=` `2` $$\frac{1}{2}\int_{1}^{3} \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$$ `=` $$\frac{1}{2}\int_{1}^{3} \frac{\color{#9a00c7}{2}}{\color{#D800AD}{2x-1}}dx$$ Substitute known values `=` `[1/2 ln (2x-1)]_1^3` Finally, get the difference of the upper and lower limits substituted to the integral as `x`.`[1/2 ln (2x-1)]_1^3` `=` `1/2[ln (2(3)-1)-ln(2(1)-1)]` Substitute the limits `=` `1/2[ln 5-0]` Evaluate `=` `1/2 ln 5` `1/2 ln 5` -
Question 4 of 6
4. Question
Find the area under the curve `y=4/x`
Round your answer to three decimal places- (4.394)`\text(units)^2`
Hint
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Fantastic!
Incorrect
Integral of a Basic Fraction
$$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$$General Form Integral of a Fraction
$$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$$First, form a fraction to balance the equation.$$\frac{4}{\color{#CC0000}{x}}$$ `=` $$\frac{4}{\color{#CC0000}{4}}$$ Differentiate `x^3+2` `=` `4` `4/1=4` We can see the limits indicated in the graph`\text(Upper Limit)` `=` `3` `\text(Lower Limit)` `=` `1` Use `4` as a constant to balance the integral.`f(x)` `=` `x` `f'(x)` `=` `1` $$4\int_{1}^{3} \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$$ `=` $$4\int_{1}^{7} \frac{\color{#9a00c7}{1}}{\color{#D800AD}{x}}dx$$ Substitute known values `=` `4[ln x]_1^3` Finally, get the difference of the upper and lower limits substituted to the integral as `x`.`4[ln x]_1^3` `=` `4[ln (3)-ln (1)]` Substitute the limits `=` `4[ln 3-ln 1]` Evaluate `=` `4 ln 3` `=` `4.394` Compute using calculator `4.394 \text(units)^2` -
Question 5 of 6
5. Question
Find the area under the curve `y=x/(x^2+1)` going to the x-axis
Round your answer to two decimal places- (0.80, 0.8)`\text(units)^2`
Hint
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Excellent!
Incorrect
Integral of a Basic Fraction
$$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$$General Form Integral of a Fraction
$$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$$First, form a fraction to balance the equation.$$\frac{x}{\color{#CC0000}{x^2+1}}$$ `=` $$\frac{x}{\color{#CC0000}{2x}}$$ Differentiate `x^3+2` `=` `1/2` `x/x=1` We can see the limits indicated in the graph`\text(Upper Limit)` `=` `2` `\text(Lower Limit)` `=` `0` Use `1/2` as a constant to balance the integral.`f(x)` `=` `x^2+1` `f'(x)` `=` `2x` $$\frac{1}{2}\int_{0}^{2} \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$$ `=` $$\frac{1}{2}\int_{0}^{2} \frac{\color{#9a00c7}{2x}}{\color{#D800AD}{x^2+1}}dx$$ Substitute known values `=` `1/2[ln (x^2+1)]_0^2` Finally, get the difference of the upper and lower limits substituted to the integral as `x`.`1/2[ln (x^2+1)]_0^2` `=` `1/2[ln ((2)^2+1)-ln ((0)^2+1)]` Substitute the limits `=` `1/2[ln 5-ln 1]` Evaluate `=` `1/2 ln 5` `=` `0.80` Compute using calculator `0.80 \text(units)^2` -
Question 6 of 6
6. Question
Find the area under the curve `y=lnx` going to the x-axis
Hint
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Nice Job!
Incorrect
Integral of a Basic Fraction
$$\int{\frac{1}{\color{#004ec4}{x}}}dx=\log_e \color{#004ec4}{x}+c$$General Form Integral of a Fraction
$$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$$Since we cannot integrate `ln x`, we can find the value of `x` and integrate it.`y` `=` `ln x` `e^y` `=` `e^(ln x)` Insert `e` as base to both sides `e^y` `=` `x` `e^(ln x)=x` `x` `=` `e^y` Therefore, the formula that we can use to find the area is `int e^y dy`We also need to find the limits for `y``\text(Lower Limit)` `=` `0` We can see on the graph that the upper limit on the x-axis is `3`Substitute this value to the equation to find the upper limit on the y-axis`y` `=` `ln x` `y` `=` `ln 3` `x=3` `\text(Upper Limit)` `=` `ln 3` Next, integrate and get the difference of the upper and lower limits substituted to the
integral as `x`.`int_0^(ln3) e^y dy` `=` `[e^y]_0^(ln3)` Integrate `=` `e^((ln3))-e^((0))` Substitute the limits `=` `3-1` Anything raised to `0` is `1` `=` `2` The area of the curve going to the y-axis is `2 \text(units)^2`To find the area of the curve going to the x-axis, find the area of the whole rectangle and subtract the area of the curve going to the y-axis.`\text(Area)` `=` `\text(length)times\text(width) ` `=` `ln 3times3` `=` `3ln 3-` `2` `3ln 3-2 \text(units)^2`
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