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1 of 8 0-day streak

Find the area under a logarithmic curve using its inverse

Question 1 of 8

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The logarithmic function and its interval are given below

g(x)=3\,\log_e\left(x\right)

The domain is \(\left(0,\,\infty\right)\), where the function is one-to-one

The input interval is

p=1,\quad q=e^{2}

Calculate the area under the curve from the displayed data

(a)1 mark

Choose the domain of \(g\) and the matching domain and range of \(g^{-1}\)

D_g=\left(-\infty,0\right),\quad R_g=\mathbb{R};\quad D_g^{-1}=\mathbb{R},\quad R_g^{-1}=\left(-\infty,0\right)

D_g=\left(0,\,\infty\right),\quad R_g=\mathbb{R};\quad D_g^{-1}=\mathbb{R},\quad R_g^{-1}=\left(0,\,\infty\right)

D_g=\mathbb{R},\quad R_g=\mathbb{R};\quad D_g^{-1}=\mathbb{R},\quad R_g^{-1}=\mathbb{R}

D_g=\left[0,\,\infty\right),\quad R_g=\mathbb{R};\quad D_g^{-1}=\mathbb{R},\quad R_g^{-1}=\left[0,\,\infty\right)

(b)1 mark

Select the inverse rule for the stated function

g^{-1}(y)=e^{\frac{4y}{3}} + 1

g^{-1}(y)=3\,\log_e\left(y\right)

g^{-1}(y)=\frac{e^{\frac{y}{3}}}{2}

g^{-1}(y)=e^{\frac{y}{3}}

g^{-1}(y)=e^{\frac{y}{3}} + 1

Complete the displayed inverse rule by entering the three missing values

g^{-1}(y)=\frac{e^{y/\square}+\square}{\square}

denominator in the exponent constant added to the exponential denominator outside the numerator

(d)1 mark

Choose the calculation that uses the inverse rule with the correct output interval

Use the inverse integral without the endpoint term

Integrate the logarithmic rule over the output interval

Use the inverse rule and integrate it between the two output values

Integrate the inverse rule between the two input values

(e)3 marks

Complete the exact area in the form \(A=Ce^{2}+D\)

coefficient of the exponential term exponent of e additive constant