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Answer 1

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In Architecture:

The eiffel tower is considered one of the best examples of the use of calculus in architecture. In order for the tower to resist wind and provide stability the tower had to be made of curves.

• To achieve these curves the architects used two derivatives for the structure

y=e^{-x}, y=-e^{-x}

• The tangent lines of the tower were used to locate the center of the tower, where its mass is focused

y=-e^(-1)x + e^(-1)(2)

y=e^(-1)x — e^(-1)(2)

Answer 2

Answer:

This is called as 'Proof of integral as a limit of sum'

Step-by-step explanation:

The definite integral of any function can be expressed either as the limit of a sum or if there exists an antiderivative F for the interval [a, b], then the definite integral of the function is the difference of the values at points a and b. Let us discuss definite integrals as a limit of a sum.

The integral of f(x) is the area of the region bounded by the curve y = f(x). This area is represented by the region ABCD as shown in the above figure. This entire region lying between [a, b] is divided into n equal subintervals given by [x₀, x₁], [x₁, x], …… [xr₋₁, xr], [xₙ₋₁, xₙ].

Let us consider the width of each subinterval as h such that h → 0, x₀ = a, x₁ = a + h, x₎ = a + 2h,…..,xr = a + rh, xₙ = b = a + nh

and n = (b – a)/h

Also, n→∞ in the above representation.

Now, from the above figure, we write the areas of particular regions and intervals as:

Area of rectangle PQFR < area of the region PQSRP < area of rectangle PQSE ….(1)

Since. h→ 0, therefore xr–  xr-1→ 0. Following sums can be established as;

$\Large {s_n = h \ [f(x_0) + ... + f(x_{n-1})] = h \sum_{r=0} ^ {n-1}f(x_r)}\\\\\Large {S_n = h \ [f(x_1) + ... + f(x_{n})] = h \sum_{r=1} ^ {n}f(x_r)}$

From the first inequality, considering any arbitrary subinterval [xr-1, xr] where r = 1, 2, 3….n, it can be said that, sₙ< area of the region ABCD <Sₙ

Since, n→∞, the rectangular strips are very narrow, it can be assumed that the limiting values of sn and Sn are equal and the common limiting value gives us the area under the curve, i.e.,

$\lim_{n \rightarrow \infty}S_n = \lim_{n \rightarrow \infty}s_n = \text{Area of the region ABCD} = \int\limits^b_a {f(x)} \, dx$

From this, it can be said that this area is also the limiting value of an area lying between the rectangles below and above the curve. Therefore,

$\Large {\int\limits^b_a {f(x)} \, dx = lim_{h \rightarrow 0} \ h [f(a) + f(a + h) + ... + f(a + (n-1)h)]}\\\\\large {\int\limits^b_a {f(x)} \, dx = (b-a) lim_{n \rightarrow \infty} \ \frac{1}{n} [f(a) + f(a + h) + ... + f(a + (n-1)h)]}\\\\\large \text {{where}},\\\\\large {h = \frac{b-a}{n} \ \ as \ n\rightarrow \infty }$$\large {\text { \therefore {\int\limits^b_a {f(x)} \, dx = lim _{n \rightarrow \infty} \sum _{i=1} ^n f(x_i) \Delta x} }}$

This is known as the definition of definite integral as the limit of sum.

Hope it helps!