#### To determine

**To estimate:**

The x-intercepts and using the intercepts, find the area of the region that lies under the curve and above the x axis.

#### Answer

2.1841

#### Explanation

**1) Concept:**

The fundamental theorem of calculus part 2: If f is continuous on a,b, then ∫abfxdx=Fb-Fa, where F is any antiderivative of f , that is, the function F such that F'=f .

**2) Formula:**

∫fx+gxdx=∫fxdx+∫gxdx

∫xndx=xn+1n+1+C

∫kdx=kx+C

**3) Given:**

y=2x+3x4-2x6

**4) Calculation:**

Consider the given curve,

y=2x+3x4-2x6

The graph of the given curve is

By using the graph of given curve, the x intercept of the given curve is x≈0, x≈1.38

Next, find the area of the region that lies under the curve and above the x axis.

From the graph, the curve y=2x+3x4-2x6 bounded between x=0 and x=1.38 is continuous on 0, 1.38 and then, the area of the region that lies under the curve and above the x axis is

∫01.38(2x+3x4-2x6)dx

To find area of region lies under the curve and above the x axis, evaluate the definite integral.

∫01.38(2x+3x4-2x6)dx

Using the addition property of integral, separate the integral.

∫01.38(2x+3x4-2x6)dx=∫01.382xdx+∫01.383x4dx-∫01.382x6dx

Applying the power rule of integral, it becomes

=2·x2201.38+3·x5501.38-2x7701.38

Using the fundamental theorem of calculus,

=1.382-0+3·1.3855-0-2·1.3877-0

By simplifying,

=2.1841 So

∫01.38(2x+3x4-2x6)dx=2.1841

**Conclusion:**

Therefore, the area of the region that lies under the given curve and above the x axis is 2.1841.