MTH108 Calculus II SUSS Assignment Sample Singapore

The MTH108 Calculus II course is the continuation of the MTH107 Calculus I course. It covers more advanced topics in calculus including integration, differentiation, and vector calculus. This course is essential for students who wish to pursue a career in mathematics, physics, engineering, or any other field that requires a strong understanding of calculus.

Assignment Outline 1: Solve derivatives of certain integrals using the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus relates differentiation and integration, and provides a powerful tool for computing derivatives of certain integrals. Specifically, if f(x) is a continuous function on the interval [a, b], and if F(x) is any antiderivative of f(x), then we have:

∫a^b f(x) dx = F(b) – F(a)

Using this theorem, we can easily compute the derivatives of certain integrals. Specifically, if we have an integral of the form:

∫g(u(x)) dx

where g(u) is a continuous function, and u(x) is a differentiable function, then we can compute its derivative as follows:

(d/dx) ∫g(u(x)) dx = g(u(x)) du/dx

To see why this is true, let F(v) be any antiderivative of g(v). Then, by the chain rule, we have:

(d/dx) F(u(x)) = F'(u(x)) du/dx

But since F(v) is an antiderivative of g(v), we have:

F'(v) = g(v)

So, we can substitute F'(u(x)) for g(u(x)), and obtain:

(d/dx) ∫g(u(x)) dx = g(u(x)) du/dx

Assignment Outline 2: Determine the derivative of certain functions.

Assignment Outline 3: Interpret limits of certain Riemann sums as definite integrals and vice versa.

Riemann sums and definite integrals are closely related concepts in calculus. A Riemann sum is an approximation of the area under a curve by dividing it into a finite number of rectangles and adding up their areas. A definite integral, on the other hand, is the exact value of the area under a curve between two given limits.

To interpret limits of certain Riemann sums as definite integrals, we can start with a function f(x) that is integrable on the interval [a, b]. Let P be a partition of [a, b], such that a = x0 < x1 < x2 < … < xn = b, and let Δxi = xi – xi-1 be the width of each subinterval. Let ti be a sample point in the i-th subinterval, so that ti belongs to [xi-1, xi]. Then the Riemann sum for f(x) over the partition P is given by:

Σf(ti)Δxi, for i = 1 to n

As we refine the partition by taking smaller subintervals, the Riemann sum converges to a limit, which is the definite integral of f(x) over [a, b]. That is,

lim n→∞ Σf(ti)Δxi = ∫a^b f(x) dx

This means that we can interpret the limit of a Riemann sum as the definite integral of the corresponding function over the given interval.

Conversely, we can also interpret a definite integral as a limit of certain Riemann sums. Specifically, we can consider the limit of Riemann sums as the width of the subintervals approaches zero. Let Pn be a partition of [a, b] with n subintervals of equal width Δx = (b – a) / n. Let ti be any point in the i-th subinterval [xi-1, xi]. Then the Riemann sum for f(x) over Pn is given by:

Σf(ti)Δx, for i = 1 to n

As n approaches infinity, the width of the subintervals approaches zero and the Riemann sum converges to the definite integral of f(x) over [a, b]. That is,

lim n→∞ Σf(ti)Δx = ∫a^b f(x) dx

Therefore, we can interpret a definite integral as the limit of Riemann sums as the partition is refined and the width of the subintervals approaches zero.

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Assignment Outline 4: Show that certain reduction formula holds.

To demonstrate that a particular reduction formula holds, we need to show that it satisfies two conditions:

1. The formula must be valid for some base case or initial value.
2. Assuming that the formula holds for some value of n, we need to show that it also holds for n + 1.

Let’s take an example to illustrate this process. Suppose we want to show that the following reduction formula holds:

∫ cos^n(x) dx = (1/n) cos^(n-1)(x) sin(x) + (n-1)/n ∫ cos^(n-2)(x) dx

To prove that this formula holds, we need to demonstrate the two conditions mentioned above:

1. Base case: For n=1, the formula reduces to:

∫ cos(x) dx = sin(x) + C

which is the well-known integral of cosine, so the formula holds for n=1.

1. Inductive step: Assume that the formula holds for some n, i.e.,

∫ cos^n(x) dx = (1/n) cos^(n-1)(x) sin(x) + (n-1)/n ∫ cos^(n-2)(x) dx

We need to show that it also holds for n+1, i.e.,

∫ cos^(n+1)(x) dx = (1/(n+1)) cos^n(x) sin(x) + n/(n+1) ∫ cos^n-1(x) dx

We start by using integration by parts, taking u=cos^n(x) and dv=dx:

∫ cos^(n+1)(x) dx = cos^n(x) sin(x) – n ∫ cos^(n-1)(x) sin^2(x) dx

We can replace sin^2(x) by 1-cos^2(x) to get:

∫ cos^(n+1)(x) dx = cos^n(x) sin(x) – n ∫ cos^(n-1)(x) (1-cos^2(x)) dx

Now, we use the assumption that the formula holds for n, to replace the integral on the right-hand side:

∫ cos^(n+1)(x) dx = cos^n(x) sin(x) – n [(n-1)/n ∫ cos^(n-2)(x) dx] + C

Simplifying the right-hand side gives:

∫ cos^(n+1)(x) dx = (1/(n+1)) cos^n(x) sin(x) + n/(n+1) ∫ cos^n-1(x) dx + C

This is the formula we wanted to prove for n+1. Therefore, we have demonstrated that the reduction formula holds for all positive integers n.

Assignment Outline 5: Use various techniques of integration to evaluate integrals.

1. Substitution: This technique involves substituting a variable in the integral with a new variable that simplifies the integral. For example, if we have the integral of (2x + 1)^2 dx, we can substitute u = 2x + 1, which gives us the integral of u^2/2 du. This integral is much easier to evaluate.
2. Integration by parts: This technique is useful for integrating products of functions. It involves choosing one function to differentiate and another function to integrate. The formula for integration by parts is ∫u dv = uv – ∫v du. For example, if we have the integral of x sin x dx, we can choose u = x and dv = sin x dx. Then, we can use integration by parts to get the integral of x sin x dx = -x cos x + sin x + C.
3. Partial fractions: This technique is used to break down a complicated fraction into simpler fractions. It is particularly useful when integrating rational functions. For example, if we have the integral of (3x + 2)/(x^2 + 3x + 2) dx, we can use partial fractions to break down the fraction into (A/(x+1)) + (B/(x+2)), where A and B are constants. Then, we can integrate each term separately.
4. Trigonometric substitutions: This technique is used to simplify integrals that involve square roots of quadratic expressions. It involves substituting a trigonometric function for the variable in the integral. For example, if we have the integral of sqrt(4x^2 + 1) dx, we can use the substitution x = (1/2)tan(u), which simplifies the integral into ∫sec(u) du.
5. Integration by partial fractions: This technique is used to simplify integrals of rational functions. It involves breaking down the function into simpler fractions and then integrating each fraction separately. For example, if we have the integral of (2x + 1)/(x^2 + x) dx, we can use partial fractions to break down the fraction into A/x + B/(x+1), where A and B are constants. Then, we can integrate each term separately.

These are just a few of the many techniques that can be used to evaluate integrals. Depending on the specific integral, one or more of these techniques may be useful.

Assignment Outline 6: Apply integration to find areas between curves or volumes of cylindrical shells.

Integration can be used to find the areas between curves or volumes of cylindrical shells. Here’s how:

Finding the area between two curves:

1. To find the area between two curves, you first need to determine the points of intersection between the two curves. Then, you can set up an integral that takes the difference between the two curves, and integrates it with respect to x or y (depending on which variable you used to graph the curves).

For example, let’s say you want to find the area between the curves y = x^2 and y = 2x – 3. To find the points of intersection, set the two equations equal to each other:

x^2 = 2x – 3

x^2 – 2x + 3 = 0

Solving for x, you get x = 1 ± √2. These are the points of intersection between the two curves. To find the area between them, you can integrate the difference between the two curves:

A = ∫[1-√2, 1+√2] (2x – 3 – x^2) dx

Evaluating this integral gives you the area between the two curves.

Finding the volume of a cylindrical shell:

1. To find the volume of a cylindrical shell, you need to integrate the area of a cross-section of the shell over the height of the shell. The cross-section of a cylindrical shell is a ring, so its area is given by A = 2πrh, where r is the radius of the shell and h is the height of the shell.

For example, let’s say you have a cylindrical shell with inner radius r1, outer radius r2, and height h. To find its volume, you can integrate the area of the cross-section over the height of the shell:

V = ∫[0, h] (2πrh) dh

Simplifying the integral gives you:

V = πh(r2^2 – r1^2)

So the volume of the cylindrical shell is πh times the difference between the square of the outer and inner radii.

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