Answer: \(\dfrac{14}{3}-2\ln3\).

The two curves meet where

\(2+5\mathrm{e}^x=4-3\mathrm{e}^{2x}\).

Let \(u=\mathrm{e}^x\). Since \(\mathrm{e}^x\gt0\), we need \(u\gt0\). The equation becomes

\(2+5u=4-3u^2\).

Rearranging,

\(3u^2+5u-2=0\).

Factorising,

\((3u-1)(u+2)=0\).

Since \(u\gt0\), we take

\(u=\frac13\).

Thus

\(\mathrm{e}^x=\frac13\), so \(x=-\ln3\).

The other vertical boundary shown is the \(y\)-axis, \(x=0\). On the interval \(-\ln3\leqslant x\leqslant0\), the curve

\(y=2+5\mathrm{e}^x\)

lies above

\(y=4-3\mathrm{e}^{2x}\).

Therefore the shaded area is

\(\displaystyle \int_{-\ln3}^{0}\left[(2+5\mathrm{e}^x)-(4-3\mathrm{e}^{2x})\right]\,\mathrm{d}x\).

Simplifying the integrand,

\((2+5\mathrm{e}^x)-(4-3\mathrm{e}^{2x})=3\mathrm{e}^{2x}+5\mathrm{e}^x-2\).

So the area is

\(\displaystyle \int_{-\ln3}^{0}(3\mathrm{e}^{2x}+5\mathrm{e}^x-2)\,\mathrm{d}x\).

An antiderivative is

\(\dfrac32\mathrm{e}^{2x}+5\mathrm{e}^x-2x\).

Substituting the limits gives

\(\left[\dfrac32\mathrm{e}^{2x}+5\mathrm{e}^x-2x\right]_{-\ln3}^{0}\).

At \(x=0\), this is

\(\dfrac32+5=\dfrac{13}{2}\).

At \(x=-\ln3\), \(\mathrm{e}^x=\frac13\) and \(\mathrm{e}^{2x}=\frac19\), so the value is

\(\dfrac32\cdot\dfrac19+5\cdot\dfrac13-2(-\ln3)=\dfrac16+\dfrac53+2\ln3=\dfrac{11}{6}+2\ln3\).

Hence the area is

\(\dfrac{13}{2}-\left(\dfrac{11}{6}+2\ln3\right)=\dfrac{14}{3}-2\ln3\).

Answer: \(\frac{96}{5}\).

Find the points of intersection by equating the two curves:

\(12-x^2=x^4-4x^2+8.\)

Rearranging,

\(x^4-3x^2-4=0.\)

Treat this as a quadratic in \(x^2\):

\((x^2-4)(x^2+1)=0.\)

The real solutions come from \(x^2-4=0\), so

\(x=-2\quad\text{or}\quad x=2.\)

Between \(-2\) and \(2\), for example at \(x=0\), the curve \(y=12-x^2\) has value \(12\), while \(y=x^4-4x^2+8\) has value \(8\). So \(12-x^2\) is above \(x^4-4x^2+8\) on the enclosed region.

The required area is

\(\displaystyle \int_{-2}^{2}\left[(12-x^2)-(x^4-4x^2+8)\right] \,\mathrm dx.\)

Simplifying the integrand gives

\(4+3x^2-x^4.\)

Therefore

\(\displaystyle \text{area}=\int_{-2}^{2}(4+3x^2-x^4)\,\mathrm dx.\)

The integrand is even, so this could also be written as twice the integral from \(0\) to \(2\). Directly integrating gives

\(\left[4x+x^3-\dfrac{x^5}{5}\right]_{-2}^{2}.\)

Substitute the limits:

\(\left(8+8-\dfrac{32}{5}\right)-\left(-8-8+\dfrac{32}{5}\right)=\dfrac{96}{5}.\)

Hence the shaded area is

\(\dfrac{96}{5}.\)

Answer: area \(=\frac43\).

The curve is

\(y=x^2-9x+18.\)

Differentiate to find the gradient of the tangent:

\(\dfrac{\mathrm dy}{\mathrm dx}=2x-9.\)

At \(x=5\),

\(\dfrac{\mathrm dy}{\mathrm dx}=2(5)-9=1.\)

Therefore the gradient of the normal is the negative reciprocal, so it is

\(-1.\)

The point on the curve where \(x=5\) has

\(y=25-45+18=-2,\)

so the point is \((5,-2)\). The equation of the normal is

\(y+2=-(x-5),\)

which simplifies to

\(y=-x+3.\)

Find where this line meets the curve:

\(x^2-9x+18=-x+3.\)

So

\(x^2-8x+15=0,\)

and hence

\((x-3)(x-5)=0.\)

The two intersections are at \(x=3\) and \(x=5\).

Between these values, for example at \(x=4\), the line has value \(-1\), while the curve has value \(-2\), so the line is above the curve. The required area is therefore

\(\displaystyle \int_3^5\left[(-x+3)-(x^2-9x+18)\right] \,\mathrm dx.\)

Simplify the integrand:

\((-x+3)-(x^2-9x+18)=-x^2+8x-15.\)

So the area is

\(\displaystyle \int_3^5(-x^2+8x-15)\,\mathrm dx =\left[-\dfrac{x^3}{3}+4x^2-15x\right]_3^5.\)

Substituting the limits gives

\(\left(-\dfrac{125}{3}+100-75\right)-\left(-9+36-45\right)=\dfrac43.\)

Therefore the area is

\(\dfrac43.\)