Verified answer

The intersection points are (0, 1) and (1, 3).

A = ∫ {0, 1} [(2x + 1) - 3^x] dx = x^2 + x - 3^x/ln 3

Evaluating: (1^2 + 1 - 3^1/ln 3) - (0^2 + 0 - 3^0/ln 3) = 2 - 3/ln 3 - 0 = 2 - 3/ln3 ≈ 0.1795

Find where the functions are equal:

3ˣ = 2x + 1

This is quite difficult to solve analytically, but by inspection, i.e. trial and error, it's fairly easy to see that x = 0 or x = 1. Check to see which function is greater on this interval by trying a value, x = 1/2 will do finely:

3^(1/2) = √3 = 1 + 1 = 2

√3 < √4 = 2

So we have that 2x + 1 > 3ˣ for 0 < x < 1. So the area between will be the integral of the difference of the functions over this interval:

A = ∫[0,1] ((2x + 1) - 3ˣ) dx

Using your integral rules this is very simple to integrate.

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After solving 3^x=2x+1 for x to find the lower lmit and upper limit of integration

ie x=0 , 1

Integrate (2x+1-3^x) between 0 and 1

x^2+x-3^x/ ln(3) evluated between 0 and 1

ie (1+1-3/ln(3))-(-1/ln(3))

2-3/ln(3)+1/ln(3)

2-2/ln(3)

2(1-1/ln(3))

If you graph these functions, they intersect at x=1 and x=3. So this will be your bounds. And 2x+1 is the "higher" function, we will subtract that from 3^x. So the answer is:

integrate from 1 to 3 of (2x+1)-(3^x)

the answer is: 11.86

Yes.