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

We need an antiderivative of \(2x + 1\).

One antiderivative for this is \(x^2 + x\). This is an antiderivative on the interval \((-\infty, \infty)\).

From theorem 5.4.3,

\(\int_1^3 (2x + 1) \mathrm{d}x = x^2 + x |_1^3 = 3^2 + 3 - (1 + 1)\)

\(= 10\)

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

We need an antiderivative of \(x^3\).

One antiderivative for this is \(\dfrac{x^4}{4}\). This is an antiderivative on the interval \((-\infty, \infty)\).

From theorem 5.4.3,

\(\int_1^4 x^3 \mathrm{d}x = \dfrac{x^4}{4} |_1^4 = 4^3 - 1 = 63\)

\(\int_{-1}^2 2x^3 - 3x^2 + 4x - 5 \mathrm{d}x\)

We need an antiderivative of \(2x^3 - 3x^2 + 4x - 5\).

One antiderivative for this is \(\dfrac{x^4}{2} - x^3 + 2x^2 - 5x\). This is an antiderivative on the interval \((-\infty, \infty)\)

From theorem 5.4.3,

\(\int_{-1}^2 2x^3 - 3x^2 + 4x - 5 \mathrm{d}x = \dfrac{x^4}{2} - x^3 + 2x^2 - 5x |_{-1}^2\)

Let's compute the remaining solution using sagemath:

x = var('x')
f = (x^4)/2 - x^3 + (2*x^2) - 5*x
f(2) - f(-1)

-21/2

\(\int_0^1 x^{100} \mathrm{d}x\)

We need an antiderivative of \(x^100\)

One antiderivative for this is \(\dfrac{x^{101}}{101}\). This is an antiderivative on the interval \((-\infty, \infty)\).

From theorem 5.4.3,

\(\int_0^1 x^{100} \mathrm{d}x = \dfrac{x^{101}}{101} |_0^1\)

Let's compute the remaining solution using sagemath:

x = var('x')
f = x^(101)/101
f(1) - f(0)

1/101

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

We need an antiderivative of \(\sqrt{x}\).

f = sqrt(x)
f
from sage.symbolic.integration.integral import indefinite_integral
g = indefinite_integral(f, x)
g

sqrt(x)
2/3*x^(3/2)

One antiderivative for this is \(\dfrac{2x^{3/2}}{3}\). This is an antiderivative on the interval \((-\infty, \infty)\)

From theorem 5.4.3,

\(\int_0^4 \sqrt{x} \mathrm{d}x = \dfrac{2x^{3/2}}{3} |_0^4\)

g
g(4) - g(0)

2/3*x^(3/2)
16/3

\(\int_{-1}^8 \sqrt[3]{x} \mathrm{d}x\)

We need an antiderivative of \(\sqrt[3]{x}\).

f = x^(1/3)
g = indefinite_integral(f, x)
g

3/4*x^(4/3)

One antiderivative for this is \(3/4*x^(4/3)\). This is an antiderivative on the interval \((-\infty, \infty)\)

From theorem 5.4.3,

\(\int_{-1}^8 \sqrt[3]{x} \mathrm{d}x = 3/4*x^(4/3) |_{-1}^8\)

g
g(8) - g(-1)

3/4*x^(4/3)
3/4*(-1)^(1/3) + 12

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

We need an antiderivative of \((1 + \sin x)\)

f = (1 + sin(x))
g = indefinite_integral(f, x)
g

x - cos(x)

One antiderivative for this is \(x - \cos x\). This is an antiderivative on the interval \((-\infty, \infty)\)

From theorem 5.4.3,

\(\int_0^{2\pi} (1 + \sin x) \mathrm{d}x = x - cos(x) |_0^{2\pi}\)

g
g(2*pi) - g(0)

x - cos(x)
2*pi

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

We need an antiderivative of \(\dfrac{x^3 + 3}{x^2}\)

f = (x^3 + 3)/x^2
g = indefinite_integral(f, x)
g

1/2*x^2 - 3/x

One antiderivative for this is \(1/2*x^2 - 3/x\). This is an antiderivative on the interval \((-\infty, 0) \cup (0, \infty)\)

From theorem 5.4.3,

\(\int_1^4 \dfrac{x^3 + 3}{x^2} \mathrm{d}x = 1/2*x^2 - 3/x |_1^4\)

g
g(4) - g(1)

1/2*x^2 - 3/x
39/4

\(\int_1^4 (\sqrt(x) + 1)^2 \mathrm{d}x\)

We need an antiderivative of \((\sqrt(x) + 1)^2\)

f = (sqrt(x) + 1)^2
g = indefinite_integral(f, x)
g

1/2*x^2 + 4/3*x^(3/2) + x

One antiderivative for this is \(1/2*x^2 + 4/3*x^(3/2) + x\). This is an antiderivative on the interval \([0, \infty)\).

From theorem 5.4.3,

\(\int_1^4 (\sqrt(x) + 1)^2 \mathrm{d}x = 1/2*x^2 + 4/3*x^(3/2) + x |_1^4\)

g
g(4) - g(1)

1/2*x^2 + 4/3*x^(3/2) + x
119/6

\(\int_1^4 \dfrac{(\sqrt{x} + 1)^2}{\sqrt{x}} \mathrm{d}x\)

We need an antiderivative of \(\dfrac{(\sqrt{x} + 1)^2}{\sqrt{x}}\)

f = (sqrt(x) + 1)^2/sqrt(x)
g = indefinite_integral(f,x)
g

2/3*(sqrt(x) + 1)^3

One antiderivative for this is \(1/2*x^2 + 4/3*x^(3/2) + x\). This is an antiderivative on the interval \((0, \infty)\)

From theorem 5.4.3,

\(\int_1^4 \dfrac{(\sqrt{x} + 1)^2}{\sqrt{x}} \mathrm{d}x = 2/3*(sqrt(x) + 1)^3 |_1^4\)

g
g(4) - g(1)

2/3*(sqrt(x) + 1)^3
38/3

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

We need an antiderivative of \(x^2(x+1)^3\)

f = x^2 * (x+1)^3
g = indefinite_integral(f,x)
g

1/6*x^6 + 3/5*x^5 + 3/4*x^4 + 1/3*x^3

One antiderivative for this is \(1/6*x^6 + 3/5*x^5 + 3/4*x^4 + 1/3*x^3\). This is an antiderivative on the interval \((-\infty, \infty)\).

From theorem 5.4.3,

\(\int_0^2 x^2(x+1)^3 \mathrm{d}x = 1/6*x^6 + 3/5*x^5 + 3/4*x^4 + 1/3*x^3 |_0^2\)

g(2) - g(0)

668/15

\(\int_{\pi/6}^{\pi/3} \csc x \cot x \mathrm{d}x\)

We know that \(\csc x = \dfrac{1}{\sin x}\) and \(\cot x = \dfrac{\cos x}{\sin x}\)

So the function is not defined when \(x = 0\) and \(x = 2\pi\).

f = csc(x) * cot(x)
g = indefinite_integral(f, x)
g

-1/sin(x)

One antiderivative for this is \(-\dfrac{1}{\sin x}\). This is an antiderivative on interval \((0, 2*\pi)\)

From theorem 5.4.3,

\(\int_{\pi/6}^{\pi/3} \csc x \cot x \mathrm{d}x = \dfrac{-1}{\sin x} |_{\pi/6}^{\pi/3}\)

g(pi/3) - g(pi/6)

-2/3*sqrt(3) + 2

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

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

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

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

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

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

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

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

\(= \sum_{i=1}^n ((1 + \dfrac{i}{n})^3 - 4 - \dfrac{4i}{n})*\dfrac{1}{n}\)

Let's compute the result using sagemath:

i,n = var('i n')
f = ((1 + i/n)^3 - 4 - (4*i/n)) * (1/n)
soln = sum(f, i, 1, n)
soln.expand()

3/2/n + 3/4/n^2 - 9/4

Now we know that \(R_n = 3/2/n + 3/4/n^2 - 9/4\)

\(\int_1^3 t^3 - 4t \mathrm{d}t = \lim_{n \to \infty} R_n = \dfrac{-9}{4}\)

\(\int_0^{\pi/2} \cos x \mathrm{d}x\)

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

\(= \dfrac{\pi/2}{n}\)

Next, the points that divided the interval \([0,\dfrac{\pi}{2}]\) into \(n\) smaller intervals are given by the formula

\(x_i = a + i \Delta x = 0 + i \dfrac{\pi}{2n} = \dfrac{i\pi}{2n}\)

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 \(x_i^{*} = x_i = \dfrac{i\pi}{2n}\)

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

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

\(= \sum_{i=1}^n \cos (\dfrac{i\pi}{2n}) * \dfrac{\pi}{2n}\)

Let's compute the result using sagemath:

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

-1/4*pi*cos(n*arctan2(sin(1/2*pi/n), cos(1/2*pi/n)) + arctan2(sin(1/2*pi/n), cos(1/2*pi/n)))*cos(1/2*pi/n)/(n*(cos(1/2*pi/n) - 1)) - 1/4*pi*sin(n*arctan2(sin(1/2*pi/n), cos(1/2*pi/n)) + arctan2(sin(1/2*pi/n), cos(1/2*pi/n)))*sin(1/2*pi/n)/(n*(cos(1/2*pi/n) - 1)) + 1/4*pi*cos(n*arctan2(sin(1/2*pi/n), cos(1/2*pi/n)) + arctan2(sin(1/2*pi/n), cos(1/2*pi/n)))/(n*(cos(1/2*pi/n) - 1)) - 1/4*pi*cos(1/2*pi/n)/(n*(cos(1/2*pi/n) - 1)) + 1/4*pi/(n*(cos(1/2*pi/n) - 1))

limit(soln, n = infinity)

1

\(\int_0^{\pi/2} \cos x \mathrm{d}x = 1\)

todo

Solution