A problem of area and tangent by definite integral Find a tangent line of the curve y = LNX in the interval (2,6), so that the area of the figure enclosed by the tangent line and x = 2, x = 6, and the curve y = LNX is the smallest

Let the tangent point be (a, LNA) and the tangent be y = (x-a) / A + LNA
s=∫[(x-a)/a+lna]dx|(2,6)-∫lnxdx|(2,6)
=x^2/2a-(1-lna)x-xlnx+∫x/xdx|(2,6)
=x^2/2a-(1-lna)x-xlnx+x|(2,6)
=-2/a+18/a-2lna+6lna+2ln2-6ln6
=16/a＋4lna＋2ln2－6ln6
For S (a) with respect to a,
s'=－16/a^2＋4/a
Let s' = 0,
a=4
So the minimum area is s (4) = 4 + 4ln4 + 2ln2-6ln6 = 4 + 10ln2-6ln6 = 0.1809

A problem of definite integral
Known power unit w = UI
U = 5
I = t^2
What is w in t (3-5)?
In addition, how to find the equivalent value of I over a period of time?
For example, if the equivalent value in 3-5 time period is 5, then w is u * I * t = 75W

Power P = UI = 5T & # 178;
Work w = ∫ (3 - > 5) PDT
=∫（3->5）5t²dt
=5/3*（125-9）
=580/3
≈193.3W

Ask a high school definite integral problem
Is 12 years Jiangxi eight school joint examination seek root sign next one minus one fourth times x square, between 2 and - 2 definite integral is 2 and - 2 have a similar f thing, followed by the formula, and then is a DX problem solving expert solution!

∫(-2→2)√(1-1/4x²)dx
This problem can't be solved by definite integral
You have to use geometry
The definite integral is the area
y=√(1-1/4x²)
y²=1-x²/4
x²/4+y²=1
Obviously, this is an ellipse
That is ∫ (- 2 → 2) √ (1-1 / 4x & # 178;) DX
It's half the area of the ellipse
The formula of elliptic area is π AB = π * 2 * 1 = 2 π
∴∫(-2→2)√(1-1/4x²)dx=2π*1/2=π

A problem of area and tangent by definite integral Find a tangent line of the curve y = LNX in the interval (2,6), so that the area of the figure enclosed by the tangent line and x = 2, x = 6, and the curve y = LNX is the smallest