Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the region bounded by the curves $y = 2^x$ and $4y - 3x = 5$ about the $x$-axis.

We locate the (conveniently integer) intersections, confirm there are no others, and apply the washer method.

Step 1: Rewrite the line

\[y = \frac{3x+5}{4}\]

Step 2: Find intersection points

At $x=1$: exponential gives $y=2$; line gives $y=\frac{3+5}{4}=2$. Match!

At $x=-1$: exponential gives $y=\frac12$; line gives $y=\frac{-3+5}{4}=\frac12$. Match!

To confirm these are the only two intersections, let $f(x)=4\cdot2^x-3x-5$; $f'(x)=4\cdot2^x\ln2-3$ has a single sign change (from negative to positive) as $x$ increases, so $f$ has exactly one local minimum and therefore at most two real zeros — exactly the two found.

Step 3: Determine which curve is above

At $x=0$: exponential gives $y=1$; line gives $y=\frac54=1.25$. The line is above the exponential on $[-1,1]$.

Step 4: Set up the washer method

\[V = \pi\int_{-1}^{1}\left[\left(\frac{3x+5}{4}\right)^2 - 2^{2x}\right]dx = \pi\int_{-1}^1\frac{(3x+5)^2}{16}\,dx - \pi\int_{-1}^1 4^x\,dx\]

Step 5: Evaluate the first integral

\[\frac{1}{16}\int_{-1}^1(9x^2+30x+25)\,dx = \frac{1}{16}\left[3x^3+15x^2+25x\right]_{-1}^1\]

At $x=1$: $3+15+25=43$. At $x=-1$: $-3+15-25=-13$.

\[\frac{1}{16}(43-(-13)) = \frac{56}{16} = \frac72\]

Step 6: Evaluate the second integral

\[\int_{-1}^1 4^x\,dx = \left[\frac{4^x}{\ln4}\right]_{-1}^1 = \frac{1}{\ln4}\left(4-\frac14\right) = \frac{15}{4\ln4} = \frac{15}{8\ln2}\]

Step 7: Calculate the volume

\[V = \pi\left(\frac72-\frac{15}{8\ln2}\right)\]

\[\boxed{V = \pi\left(\frac72-\frac{15}{8\ln2}\right) \text{ cubic units}}\]

Step 8: Numerical value

$\dfrac{15}{8\ln2}\approx2.708$, so $V\approx\pi(3.5-2.708) = 0.792\pi \approx 2.49$ cubic units.

Summary

\[\boxed{V = \pi\left(\frac72-\frac{15}{8\ln2}\right) \approx 2.49 \text{ cubic units}}\]

Key insight: The two curves happen to intersect at the tidy integer-coordinate points $(-1,\frac12)$ and $(1,2)$, but the resulting volume still mixes a rational term from the line with a $\ln2$ term from the exponential — a reminder that clean intersection points don't guarantee a clean final answer.