Find areas involving reciprocal curves

The reciprocal curve is above the x-axis on the interval. Which setup gives its geometric area?

Move the probe across [0,4]. Compare the solid reciprocal curve with the dashed line. The graph suggests where their order changes; factorisation will prove it

How many intervals need their own upper-minus-lower expression?

Find area with a reciprocal curve

A reciprocal curve can have a singularity and may lie above or below the horizontal axis. Check the interval and sign before integrating, then convert signed values to geometric area

1. 1Locate the singularity Set the denominator equal to zero and check whether the interval contains that input value \(\displaystyle x+1=0\Rightarrow x=-1,\quad [0,3]\text{ does not contain }-1\)
2. 2Check the sign Determine whether the curve is above or below the horizontal axis on each part of the interval \(\displaystyle \frac{2}{x+1}>0\text{ on }[0,3]\Rightarrow A=\int_0^3\frac{2}{x+1}\,dx\)
3. 3Use the logarithmic primitive A reciprocal linear function integrates to a logarithm with the inner coefficient divided out \(\displaystyle \int\frac{2}{x+1}\,dx=2\log_e\left|x+1\right|+C\)
4. 4Split when the upper curve changes If another curve crosses the reciprocal curve, split at every intersection and use upper minus lower on each part \(\displaystyle A=\int_0^1\left(\frac{2}{x+1}-1\right)\,dx+\int_1^3\left(1-\frac{2}{x+1}\right)\,dx\)

Curve data

\(\displaystyle x+1=0\Rightarrow x=\square\)

Find the input value of the vertical asymptote

Curve data

\(\displaystyle \frac{3+1}{0+1}=\square\)

Combine the endpoint inner values to obtain the positive logarithm argument

Negating an above-axis integral. One of these lines is right. Which one?

Use distances from the actual singularity

For this curve the singularity is x=2 and the whole interval is to its left. The curve is negative there. Factor the inner coefficient out when finding a primitive, then negate the signed integral for geometric area. The endpoint distances from x=2 are 2 and 1; reflection is about this singularity, not automatically about the y-axis

Curve and interval

\(\displaystyle f(x)=\frac{9}{3x-6},\quad [0,1],\quad A=K\ln(q),\quad K=|9/3|\)

Find the area coefficient shown below. Keep it equal to the magnitude of the reciprocal primitive coefficient

Curve and interval

\(\displaystyle f(x)=\frac{9}{3x-6},\quad [0,1],\quad A=3\ln(q)\)

Find q>1 after converting the signed integral to positive area. Keep K unchanged

Separate signed integral from area

The primitive is valid throughout [0,1], which does not contain the singularity. Its change is negative, but geometric area is positive

1. 1Evaluate the signed integral Use absolute values inside the real logarithm \(\displaystyle \int_0^1\frac9{3x-6}\,dx=3\ln3-3\ln6=-3\ln2\)
2. 2Convert to geometric area The negative branch requires the opposite of its signed integral \(\displaystyle A=3\ln2\)
3. 3Check by distance substitution With t=2−x, the endpoint distances run from 2 down to 1. The two sign changes give the same positive area \(\displaystyle A=\int_0^1\frac3{2-x}\,dx=\int_1^2\frac3t\,dt=3\ln2\)

The entry form fixes K=3 and requires q>1 in K ln(q). Which pair obeys that convention?

Locate every change in curve order

Find both intersections before deciding which curve is higher. The denominator is positive throughout [0,4], so clearing it introduces no excluded root within this interval

Curve and interval

\(\displaystyle \frac{12}{x+1}=\frac{15-3x}{2},\quad 0\le x\le4\)

Solve f(x)=g(x) and enter the smaller intersection input

Curve and interval

\(\displaystyle \frac{12}{x+1}=\frac{15-3x}{2},\quad 0\le x\le4\)

Enter the larger intersection input

Prove the three signs

The denominator is positive. The two numerator factors have the same sign before 1 and after 3, but opposite signs between 1 and 3. Thus f−g has signs +, −, + on the three open pieces

Which setup gives the whole geometric area on [0,4]?

Curve and interval

\(\displaystyle P(x)=12\ln(x+1)-\frac{15x}{2}+\frac{3x^2}{4},\quad A=2P(1)-2P(3)+P(4)-P(0)\)

Evaluate the three-piece area and find q in A=12 ln(q). Keep the coefficient 12 fixed

Combine all three changes of the primitive

P is a primitive of f−g on the whole interval. Negate its change only on the middle piece, where g is higher

1. 1Use the signs Collect the four endpoint values after treating each piece separately \(\displaystyle A=[P(1)-P(0)]-[P(3)-P(1)]+[P(4)-P(3)]\)
2. 2Combine logarithms The rational terms cancel in this example. That cancellation is not a general rule for area between reciprocal curves and lines \(\displaystyle A=12(2\ln2-2\ln4+\ln5)=12\ln(5/4)\)
3. 3Check the sign and entry form The argument 5/4 exceeds one and the reciprocal coefficient stays 12. \(\displaystyle K=12,\quad q=\frac54,\quad A>0\)

Construct and justify the complete area, ungraded

Sketch both curves on the given interval. Locate the singularity, solve both intersections, prove the curve order on all three pieces and derive the exact geometric area. Write the full setup and working on paper

Curve and interval

\(\displaystyle f(x)=\frac6x,\quad g(x)=5-x,\quad \left[\frac32,\frac72\right],\quad A=6\ln(q),\quad q>1\)

After completing the derivation, enter the positive logarithm argument shown below, keeping the coefficient six fixed

Check the full three-piece argument

The singularity x=0 is outside the interval. Compare every piece of your written derivation, not only its final value

1. 1Prove intersections and order The denominator is positive; the numerator changes sign at 2 and 3, giving +, −, + \(\displaystyle f(x)-g(x)=\frac{(x-2)(x-3)}x=0\Rightarrow x=2,3\)
2. 2Use a valid primitive The logarithm is real on the whole positive interval \(\displaystyle Q(x)=6\ln x-5x+\frac{x^2}{2}\)
3. 3Write three upper-minus-lower integrals Use the proved curve order on each piece \(\displaystyle A=\int_{3/2}^{2}(f-g)\,dx+\int_2^3(g-f)\,dx+\int_3^{7/2}(f-g)\,dx\)
4. 4Combine three geometric pieces The middle change is negated. In this example the rational terms again cancel \(\displaystyle A=2Q(2)-2Q(3)+Q(7/2)-Q(3/2)=2(6\ln2-8)-2(6\ln3-\frac{21}{2})+(6\ln\frac72-\frac{91}{8})-(6\ln\frac32-\frac{51}{8})=6\ln\frac{28}{27}\)

Can ordinary endpoint subtraction find the area under this curve across the stated interval?

Check the domain before adding geometric pieces

Exclude singularities from each ordinary integration interval, prove where the upper curve changes and use upper minus lower on every piece. For coefficient entries, keep the specified reciprocal coefficient and a logarithm argument greater than one