The problem of finding the original function in calculus, Ax ^ 2 / 1 + A ^ 2 + x ^ 2 (a is constant) The numerator is ax ^ 2 and the denominator is 1 + A ^ 2 + x ^ 2 Find primitive function The original question should be ∫ (AX ^ 2 / 1 + A ^ 2 + x ^ 2) DX

The problem of finding the original function in calculus, Ax ^ 2 / 1 + A ^ 2 + x ^ 2 (a is constant) The numerator is ax ^ 2 and the denominator is 1 + A ^ 2 + x ^ 2 Find primitive function The original question should be ∫ (AX ^ 2 / 1 + A ^ 2 + x ^ 2) DX

Tip: first simplify the numerator and denominator of ax ^ 2 / 1 + A ^ 2 + x ^ 2aX ^ 2 / 1 + A ^ 2 + x ^ 2 = a (x ^ 2 / 1 + A ^ 2 + x ^ 2) = a (1 + A ^ 2 + x ^ 2-1-a ^ 2) / (1 + A ^ 2 + x ^ 2) = a [1 - (1 + A ^ 2) / (1 + A ^ 2 + x ^ 2)] (1 + A ^ 2) / (1 + A ^ 2 + x ^ 2) divided by 1 + A ^ 2 can be reduced to 1 / [1 + (x / (√ 1 + A ^ 2)) ^ 2]

Calculus function There is such a question: A certain reservoir continuously flows in a certain amount of water every day. According to the current discharge capacity, the water in the reservoir can be used for 80 days. However, due to the recent hot weather, the inflow is reduced by 20%. If the discharge capacity remains unchanged, it can only be used for 60 days. If it still needs to be used for 80 days, what percentage should the discharge capacity be reduced every day? (this topic is calculated by a binary primary equation. Now can we turn it into a function problem calculated by definite integral?) (it's best to calculate by definite integral in senior high school)

Originally, let the velocity function of water inflow be V1 (T) and water outflow be V2 (T). Let's know that ∫ 0-80 {(V1 (T) - V2 (T)} DX = C ∫ 0-60 {80% V1 (T) - V2 (T) DX = C. calculate the integral 0-80 {80% V1 (T) - Kv2 (T) DX = C. The total water quantity can be solved by solving this equation.. in the University, V1 (T) V2 (T) are all in the real domain

The problem of finding primitive function by calculus Find the original function of T / (1 + cost)

∫[t/(1+cost)]dt=∫[t(1-cost)/sin ² t]dt
=∫[t/sin ² t]dt-∫[tcost/sin ² t]dt
=∫tcsc ² tdt-∫[tcost/sin ² t]dt
From the first integral:
∫tcsc ² tdt=-∫td(cott)=-[tcott-∫cottdt]
=-tcott+∫cottdt=-tcott+ln(sint)
From the second integral:
∫[tcost/sin ² t]dt=-∫td(1/sint)=-t/sint+∫(dt/sint)
=-t/sint+ln|csct-cott|
Finally:
∫[t/(1+cost)]dt=-tcott+ln(sint)+t/sint-ln|csct-cott|+c

Calculus function proof Let f (x) be any function defined in (- A, a). It is proved that f (x) can always be expressed as the sum of even function and odd function I'm a beginner in calculus,

Let f (x) be any function defined in (- A, a), let: H (x) = [f (x) + F (- x)] / 2, G (x) = [f (x) - f (- x)] / 2, then H (x) + G (x) = f (x), and H (- x) = [f (- x) + F (x)] / 2 = [f (x) + F (- x)] / 2 = H (x) is an even function, G (- x) = [f (- x) - f (x)] / 2 = - [f (x) - f (- x)] / 2 = - G (x) is an odd function, so

How to calculate the area of the enclosed figure by calculus Y = - x ^ 2 + 1 and the area of the figure surrounded by the X axis

When y = 0, x = 1 or - 1, the integral of Y is - (1 / 3) x ^ 3 + X + C Y (x = - 1) = - 2 / 3 + C Y (x = 1) = 2 / 3 + C S = y (x = 1) - Y (x = - 1) = 4 / 3, which is the area enclosed by the curve and X axis

The problem of calculating graphic area by calculus Thank you for your advice 1. Find the area enclosed by y = LNX, x = 0 and the straight line y = LNA, y = LNB (b > a > 0) 2. Find the area of the graph surrounded by y = e ^ x, y = e ^ - X and x = 1. (x > 0) 3. Try to find the area of the plane figure enclosed by the curve y ^ 2 = 2x + 1 and y = X-1

1 y=lnx =>x=e^y
The area is the area between X = e ^ y and the Y axis. The upper limit of the integral is x = B, the lower limit is x = a, and the answer is e ^ B-E ^ a
2 ∫ (e ^ x-e ^-x) DX 0-1 result e + 1 / E-2
3 ∫ (y + 1) - (y ^ 2 / 2-1 / 2) dy - 1 ~ 3 result 16 / 3