\(\int_0^3 (2x+1) \mathrm{d}x\)

Suppose \(n\) is a positive integer. We begin by computing all of the quantities mentioned in Definition 5.3.1. We have \(f(x) = 2x + 1, a = 0, b = 3\), so \(\Delta x = \dfrac{(b-a)}{n}\)

\(= \dfrac{3-0}{n} = \dfrac{3}{n}\)

Next, the points that divided the interval \([0,3]\) into \(n\) smaller intervals are given by the formula

\(x_i = a + i \Delta x = 0 + i \dfrac{3}{n} = \dfrac{3i}{n}\)

Finally, we must choose the sample points \(x^{*}_i \in [x_{i-1}, x_i]\). According to theorem 5.3.2, it doesn't matter what choice we make; all choices will lead to the same value when we take the limit of the Riemann \(n\) sum as \(n \to \infty\).

Let \(x_i^{*} = x_i = \dfrac{3i}{n}\)

We can now work out a formula for the Riemann sum:

\(R_n = \sum_{i=1}^n f(x_i^{*})\Delta x = \sum_{i=1}^n f(\dfrac{3i}{n}) * \dfrac{3}{n}\)

\(= \sum_{i=1}^n ( \dfrac{6i}{n} + 1)* \dfrac{3}{n}\)

Let's compute the result via SageMath:

i,n = var('i n')
f = (6*i/n + 1) * (3/n)
soln = sum(f, i, 1, n)
soln.simplify_full()

3*(4*n + 3)/n

soln.expand()

9/n + 12

Now we know that \(R_n = \dfrac{9}{n} + 12\)

\(\int_0^3 (2x+1) \mathrm{d}x = \lim_{n \to \infty} R_n = \lim_{n \to \infty} \dfrac{9}{n} + 12 = 12\)

\(\int_1^3 (2x+1) \mathrm{d}x\)

Suppose \(n\) is a positive integer. We begin by computing all of he quantities mentioned in Definition 5.3.1. We have \(f(x) = 2x + 1, a = 1, b = 3\), so \(\Delta x = \dfrac{(b-a)}{n}\)

\(= \dfrac{3-1}{n} = \dfrac{2}{n}\)

Next, the points that divided the interval \([1,3]\) into \(n\) smaller intervals are given by the formula

\(x_i = a + i \Delta x = 1 + i \dfrac{2}{n} = 1 + \dfrac{2i}{n}\)

Finally, we must choose the sample points \(x^{*}_i \in [x_{i-1}, x_i]\). According to theorem 5.3.2, it doesn't matter what choice we make; all choices will lead to the same value when we take the limit of the Riemann \(n-\) sum as \(n \to \infty\).

Let \(x_i^{*} = x_i = 1 + \dfrac{2i}{n}\)

We can now work out a formula for the Riemann sum:

\(R_n = \sum_{i=1}^n f(x_i^{*})\Delta x = \sum_{i=1}^n f(1 + \dfrac{2i}{n}) * \dfrac{2}{n}\)

\(= \sum_{i=1}^n ( 2 + \dfrac{4i}{n} + 1)* \dfrac{2}{n}\)

Let's compute the result via SageMath:

i,n = var('i n')
f = (2 + (4*i/n) + 1) * (2/n)
soln = sum(f, i, 1, n)
soln.expand()

4/n + 10

Now we know that \(R_n = \dfrac{4}{n} + 10\)

\(\int_0^3 (2x+1) \mathrm{d}x = \lim_{n \to \infty} R_n = \lim_{n \to \infty} \dfrac{4}{n} + 10 = 10\)

We can also verify the above result using sagemath:

integrate(2*x + 1, x, 1, 3)

10

\(\int_0^4 x^2 \mathrm{d}x\)

Suppose \(n\) is a positive integer. We begin by computing all of the quantities mentioned in Definition 5.3.1. We have \(f(x) = x^2, a = 0, b = 4\), so \(\Delta x = \dfrac{(b-a)}{n}\)

\(= \dfrac{4-0}{n} = \dfrac{4}{n}\)

Next, the points that divided the interval \([0,4]\) into \(n\) smaller intervals are given by the formula

\(x_i = a + i \Delta x = 0 + i \dfrac{4}{n} = \dfrac{4i}{n}\)

Finally, we must choose the sample points \(x^{*}_i \in [x_{i-1}, x_i]\). According to theorem 5.3.2, it doesn't matter what choice we make; all choices will lead to the same value when we take the limit of the Riemann \(n\) sum as \(n \to \infty\).

Let \(x_i^{*} = x_i = \dfrac{4i}{n}\)

We can now work out a formula for the Riemann sum:

\(R_n = \sum_{i=1}^n f(x_i^{*})\Delta x = \sum_{i=1}^n f(\dfrac{4i}{n}) * \dfrac{4}{n}\)

\(= \sum_{i=1}^n ( \dfrac{16i^2}{n^2})* \dfrac{4}{n}\)

Let's compute the result via SageMath:

i,n = var('i n')
f = (16 * i^2 / n^2) * (4/n)
soln = sum(f, i, 1, n)
soln.expand()

32/n + 32/3/n^2 + 64/3

So, \(R_n = 32/n + 32/3/n^2 + 64/3\)

\(\int_0^4 x^2 \mathrm{d}x = \lim_{n \to \infty} R_n = \dfrac{64}{3}\)

\(\int_1^3 (4-x^2) \mathrm{d}x\)

Suppose \(n\) is a positive integer. We begin by computing all of the quantities mentioned in Definition 5.3.1. We have \(f(x) = 4 - x^2, a = 1, b = 3\), so \(\Delta x = \dfrac{(b-a)}{n}\)

\(= \dfrac{3-1}{n} = \dfrac{2}{n}\)

Next, the points that divided the interval \([1,3]\) into \(n\) smaller intervals are given by the formula

\(x_i = a + i \Delta x = 1 + i \dfrac{2}{n} = 1 + \dfrac{2i}{n}\)

Finally, we must choose the sample points \(x^{*}_i \in [x_{i-1}, x_i]\). According to theorem 5.3.2, it doesn't matter what choice we make; all choices will lead to the same value when we take the limit of the Riemann \(n\) sum as \(n \to \infty\).

Let \(x_i^{*} = x_i = 1 + \dfrac{2i}{n}\)

We can now work out a formula for the Riemann sum:

\(R_n = \sum_{i=1}^n f(x_i^{*})\Delta x = \sum_{i=1}^n f(1 + \dfrac{2i}{n}) * \dfrac{2}{n}\)

\(= \sum_{i=1}^n ( 4 - (1 + \dfrac{2i}{n})^2)* \dfrac{2}{n}\)

Let's compute the result via SageMath:

i,n = var('i n')
f = (4 - (1 + (2*i/n))^2) * (2/n)
soln = sum(f, i, 1, n)
soln.expand()

-8/n - 4/3/n^2 - 2/3

So, \(R_n = -8/n - 4/3/n^2 - 2/3\)

\(\int_1^3 (4-x^2) \mathrm{d}x = \lim_{n \to \infty} R_n = \dfrac{-2}{3}\)

We can also verify the above result using sagemath:

integrate(4-x^2, x, 1, 3)

-2/3

\(\int_{-1}^1 (x^2 - 3x + 1) \mathrm{d}x\)

Suppose \(n\) is a positive integer. We begin by computing all of the quantities mentioned in Definition 5.3.1. We have \(f(x) = x^2 - 3x + 1, a = -1, b = 1\), so \(\Delta x = \dfrac{(b-a)}{n}\)

\(= \dfrac{1-(-1)}{n} = \dfrac{2}{n}\)

Next, the points that divided the interval \([-1,1]\) into \(n\) smaller intervals are given by the formula

\(x_i = a + i \Delta x = -1 + i \dfrac{2}{n} = \dfrac{2i}{n} - 1\)

Finally, we must choose the sample points \(x^{*}_i \in [x_{i-1}, x_i]\). According to theorem 5.3.2, it doesn't matter what choice we make; all choices will lead to the same value when we take the limit of the Riemann \(n\) sum as \(n \to \infty\).

Let \(x_i^{*} = x_i = \dfrac{2i}{n} - 1\)

We can now work out a formula for the Riemann sum:

\(R_n = \sum_{i=1}^n f(x_i^{*})\Delta x = \sum_{i=1}^n f(\dfrac{2i}{n} - 1) * \dfrac{2}{n}\)

\(= \sum_{i=1}^n ((2i/n - 1)^2 - (6i/n - 3) + 1)* \dfrac{2}{n}\)

Let's compute the result via SageMath:

i,n = var('i n')
f = (((2*i/n) - 1)^2 - ((6*i/n)-3) + 1)*(2/n)
soln = sum(f,i,1,n)
soln.expand()

-6/n + 4/3/n^2 + 8/3

So, \(R_n = -6/n + 4/3/n^2 + 8/3\)

\(\int_{-1}^1 (x^2 - 3x + 1) \mathrm{d}x = \dfrac{8}{3}\)

\(\int_1^4 x^3 \mathrm{d}x\)

Suppose \(n\) is a positive integer. We begin by computing all of the quantities mentioned in Definition 5.3.1. We have \(f(x) = x^3, a = 1, b = 4\), so \(\Delta x = \dfrac{(b-a)}{n}\)

\(= \dfrac{4-1}{n} = \dfrac{3}{n}\)

Next, the points that divided the interval \([1,4]\) into \(n\) smaller intervals are given by the formula

\(x_i = a + i \Delta x = 1 + i \dfrac{3}{n} = 1 + \dfrac{3i}{n}\)

Finally, we must choose the sample points \(x^{*}_i \in [x_{i-1}, x_i]\). According to theorem 5.3.2, it doesn't matter what choice we make; all choices will lead to the same value when we take the limit of the Riemann \(n\) sum as \(n \to \infty\).

Let \(x_i^{*} = x_i = 1 + \dfrac{3i}{n}\)

We can now work out a formula for the Riemann sum:

\(R_n = \sum_{i=1}^n f(x_i^{*})\Delta x = \sum_{i=1}^n f(1 + \dfrac{3i}{n}) * \dfrac{3}{n}\)

\(= \sum_{i=1}^n ( (1 + 3i/n)^3)* \dfrac{3}{n}\)

Let's compute the result via SageMath:

i,n = var('i n')
f = (1 + (3*i/n))^3 * (3/n)
soln = sum(f, i, 1, n)
soln.expand()

189/2/n + 135/4/n^2 + 255/4

Now we know that \(R_n = 3/4*(85*n^2 + 126*n + 45)/n^2\)

\(\int_1^4 x^3 \mathrm{d}x = \lim_{n \to \infty} R_n = \lim_{n \to \infty} 189/2/n + 135/4/n^2 + 255/4 = 255/4\)

We can also verify the above result using sagemath:

integrate(x^3, x, 1, 4)

255/4

\(\int_0^6 (2 - \dfrac{x}{2}) \mathrm{d}x\)

Suppose \(n\) is a positive integer. We begin by computing all of the quantities mentioned in Definition 5.3.1. We have \(f(x) = 2 - \dfrac{x}{2}, a = 0, b = 6\), so \(\Delta x = \dfrac{(b-a)}{n}\)

\(= \dfrac{6-0}{n} = \dfrac{6}{n}\)

Next, the points that divided the interval \([0,6]\) into \(n\) smaller intervals are given by the formula

\(x_i = a + i \Delta x = 0 + i \dfrac{6}{n} = \dfrac{6i}{n}\)

But we are asked to choose a different \(x_i\)

\(x_i = x_{i-1} = \dfrac{6(i-1)}{n}\)

According to theorem 5.3.2, it doesn't matter what choice we make for \(x_i\); all choices will lead to the same value when we take the limit of the Riemann \(n\) sum as \(n \to \infty\).

Let \(x_i^{*} = x_i = \dfrac{6(i-1)}{n}\)

We can now work out a formula for the Riemann sum:

\(R_n = \sum_{i=1}^n f(x_i^{*})\Delta x = \sum_{i=1}^n f(\dfrac{6(i-1)}{n}) * \dfrac{6}{n}\)

$= ∑_i=1ⁿ ( 2 - \dfrac{3(i-1)}{n} )* \dfrac{6}{n} $

Let's compute the result via SageMath:

i,n = var('i n')
f = (2 - ((3*(i-1))/n)) * (6/n)
soln = sum(f, i, 1, n)
soln.expand()

9/n + 3

Now we know that \(R_n = \dfrac{9}{n} + 3\)

\(\int_0^6 (2 - \dfrac{x}{2}) \mathrm{d}x = \lim_{n \to \infty} R_n = \lim_{n \to \infty} \dfrac{9}{n} + 3 = 3\)

\(\forall a,b,c \in R\)

Suppose \(a < b\)

We have \(\int_a^b c \mathrm{d}x\)

Since \(c\) is a constant, we know that \(f(x) = 1\) which is a constant function. So,

\(\int_a^b c f(x) \mathrm{d}x\)

From theorem 5.3.6,

\(c \int_a^b f(x) \mathrm{d}x\)

Since the function is constant, we know that it's minimum and maximum value are the same. So, From theorem 5.3.7,

\(m(b-a) = \int_a^b f(x) \mathrm{d}x = M(b-a)\)

Since \(m = M = 1\),

\((b-a) = \int_a^b f(x) \mathrm{d}x = (b-a)\)

Multiplying \(c\) with it,

\(c \int_a^b f(x) \mathrm{d}x = c(b-a)\)

\(\int_a^b cf(x) \mathrm{d}x = c(b-a)\)

\(\int_a^b c \mathrm{d}x = c(b-a)\)

Suppose \(a>b\) and \(f\) is continous on \([b,a]\) and let \(m\) and \(M\) be the minimum and maximum values of \(f\) on \([b,a]\).

We need to prove that \(M(b-a) \leq \int_a^b f(x) \mathrm{d}x \leq m(b-a)\)

For any positive integer \(n\), let

\(R_n = \sum_{i=1}^n f(x_i^{*}) \Delta x\)

be a Riemann \(n\) sum for \(f\) where all notation is as in Definition 5.3.1. Since \(M\) is the maximum value of \(f\) on \([b,a]\), for every \(i\) we have \(f(x_i^{*}) \leq M\), and therefore

\(R_n = \sum_{i=1}^n f(x_i^{*}) \Delta x \leq \sum_{i=1}^n M\delta x = M \delta x .n = M\dfrac{a-b}{n} = M(a-b)\)

Taking the limit as \(n \to \infty\) (and applying the preservation of weak inequalities by limits), we conclude that

\(\int_b^a f(x) \mathrm{d}x = \lim_{n \to \infty} R_n \leq M(a-b)\)

\(\int_b^a f(x) \mathrm{d}x \leq M(a-b)\)

From Definition 5.3.10,

\(- \int_a^b f(x) \mathrm{d}x \leq M(a-b)\)

\(\int_a^b f(x) \mathrm{d}x \geq M(b-a)\)

\(M(b-a) \leq \int_a^b f(x) \mathrm{d}x\)

Similar reasoning establishes the other inequality in the theorem.

I will just prove the first part of the theorem.

\(\int_a^b c f(x) \mathrm{d}x = c \int_a^b f(x) \mathrm{d}x\)

We already know from Theorem 5.3.6, that the above is true when \(a < b\). So, all we have to do is prove for \(b > a\).

Suppose \(b > a\)

For any positive integer \(n\), as usual we let \(\Delta x = \dfrac{b-a}{n}\) and \(x_i = a + i\Delta x\) and we choose sample points \(x_i^{*} \in [x_{i-1}, x_i]\). Using these sample points, we get the following Riemann summ for \(f\) on \([a,b]\):

\(R_n = \sum_{i=1}^n f(x_i^{*}) \Delta x\)

We can also use these sample points to compute a Riemann sum for \(f\):

\(\sum_{i=1}^n f(x_i^{*}) \Delta x = R_n\)

From the definition of definite integrals,

\(\int_a^b f(x) \mathrm{d}x = \lim_{n \to \infty} R_n\)

\(c \int_a^b f(x) \mathrm{d}x = c\lim_{n \to \infty} R_n\)

\(c \int_a^b f(x) \mathrm{d}x = \lim_{n \to \infty} c R_n\)

\(c \int_a^b f(x) \mathrm{d}x = \int_a^b c f(x) \mathrm{d}x\)