The answers in the book are omitted

The simplest is according to the geometric meaning.

f(x)>0 illustrates that the function is over the interval [a,b], and the image is above the x-axis.

f (x)<0 illustrates that the function is monotonically decreasing over the interval [a,b].

f (x)>0 indicates that the function is concave in the interval [a,b].

Based on the above three pieces of information, you can draw a sketch of f(x).

S1 is the area of the curved trapezoidal ABCD S3 is the trapezoidal ABCD (I forgot to connect the CD on the diagram!). The area of S2 is the area of the rectangular ABCE.

Obviously, there is s2, and the other is the comparative property of using definite integrals.

The equation for the line segment cd: y = f(a) + (x-a)[f(b)-f(a)] (b-a).

The equation for the concave arc cd: y=f(x).

The equation for the line segment CE: y=f(b).

According to the position relationship between the upper line segment CD, concave arc CD and line segment CE in the sketch, it is easy to know:

f(a)+(x-a)[f(b)-f(a)]/(b-a)>f(x) >f(b)

Integral: [a,b]dx> [a,b]f(x)dx> [a,b]f(b)dx

f(a)+f(b)](b-a)/2>∫[a,b]f(x)dx>f(b)(b-a)

i.e. s3>s1>s2

f'(x) <0 So f(b) s2 and s3 are really just a matter of comparing the sizes of 2f(b) and f(a) +f(b) So s2 < s3

S1 remains, and S3 would be nice to compare qualitatively a little.

f''(x) >0 shows that f(x) is a concave function similar to x 2, e x.

S3 represents the area of a trapezoid, and compared to S1, the top, bottom, bottom, and right-angled edges coincide.

The only difference is the hypotenuse Since f(x) is a concave function, the function image is below the hypotenuse of s3.

Therefore s3 > s1 > s2

The result in the first Sengui box, f(x)=(1+x)e -xe -x+xe -x

f'(x)=-e -x+e -x-xe -x. f'(x)=-xe^-x…Shizi....①

The conditions in the second box, looking at the formula, are first specified under this letter.

e -x > 0, we know that when x<0, f'(x)>0；

When x>0, f'(x)<0。

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Question 2. Just use it for 2, 4

2, 4 of the second question

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Hello classmates. The meaning of the integral represents the area of the siege with the x-axis.

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When the integration interval is the same, the size of the integrated function is compared.

The magnitude of the definite integral value can be determined.

The resolution is as follows: aIn [1,内e], ln x ln x, so ln xdx lnxdx does not hold.

b.[e,e], ln x ln x, so ln xdx lnxdx holds.

c.In [1,+, x x, so x dx allows x dx to hold.

d.x 4dx> x dx does not hold.

My answer is: BC

1, on (0,)], there is obviously an x definite integral.

There are inequality properties, too

xdxx^2dx.

2, on (0, 2), there is x>sinx, by the inequality property of the pure fraction of the fixed product, there is also

xdxsinxdx.

The inequality of definite integrals: if the banquet is split f(x)f(x)dx

a^b_g(x)dx

First of all, I would like to state that I have been away from university for a while, and I have been doing this question from memory.

Problem solving idea: Use the geometric meaning of the definite integral to solve the problem.

In the interval [0, 6], the image of Tanx 2 is larger than the image of Sinx 2, which represents a larger area.

Therefore: i1 so: choose a

Since the integration interval is the same, just compare the size of sinx 2 and tanx 2 in this interval.

And in this range.

sinx^2x^2

So... sinx^2tanx^2

So choose A

The nature of definite integrals is used, and it is more appropriate to fill in ">", and the specific process is referred to the figure below.