1 of 8 0-day streak

Find areas involving reciprocal curves

Question 1 of 8

No calculator

A reciprocal curve is given on an interval beside the x-axis

r(x)=\frac{\frac{3}{2}}{\left(-3\,x + 2\right)}

The interval is \(I\)=\(\left[-1,\;0\right]\)

The reciprocal curve is above the x-axis on this interval

Find the exact area between the curve and the x-axis

(a)1 mark

Find the input value of the vertical asymptote

Input value of vertical asymptote

Format: Give an exact value

(b)1 mark

Select the integral setup for the geometric area

A=\int_{-1}^{0}\frac{\frac{3}{2}}{\left(-3\,x + 2\right)}\,dx

A=\int_{0}^{-1}\frac{\frac{3}{2}}{\left(-3\,x + 2\right)}\,dx

A=-\int_{-1}^{0}\frac{\frac{3}{2}}{\left(-3\,x + 2\right)}\,dx

Enter the coefficient and positive argument in \(A=c\log_e(q)\)

Use the absolute value of the reciprocal primitive coefficient obtained directly from the displayed fraction, with its numerator and denominator exactly as printed. Combine the endpoint logarithms into one logarithm and choose q greater than 1. Do not move factors into or out of the logarithm. Enter exact values

Coefficient c Positive argument q