Derivatives and Integrals

This page covers three foundational topics:

Introduction to Derivatives
Introduction to Integrals
Practice Problems

External Resources

Book https://rodrigopacios.github.io/mrpacios/download/Thomas_Calculus.pdf

Video https://www.khanacademy.org/math/ap-calculus-ab

Source

This tutorial, the figures and the questions are translated from a Chinese Calculus text Book: 高等数学，第七版，上册，同济大学数学系编

1. Introduction to Derivatives

Velocity of Linear Motion

Let's assume an object is moving along a straight line. Once the origin, direction, and unit of length are defined, we regard the straight line as the x-axis. Let the object's position at time t be denoted as s, measured from the origin in the positive direction. Additionally, if the position at time t is measured using a measuring instrument, we obtain a time-position pair (t, s), where s is a function of t: $$ s = f(t) $$

This means the motion of the object is determined by a function.

The change in the position of the object over time is called displacement, which is a vector quantity. In simple terms, the ratio of displacement over time is called average velocity. That is, for a short time period, regardless of how we choose the starting and ending time points, the ratio:

$$Distance \; Traveled \; ÷ Time \; Used$$

is called average velocity.

This average velocity represents how fast the object moves during the interval. If the object is not moving uniformly, then the average velocity across different time intervals may differ. However, for very short intervals, if we compute this ratio in the limit as the time interval shrinks to 0, we get the instantaneous velocity at that point. Let t → t₀, and if the following limit exists:

$$ \lim_{t \to t_0} \frac{f(t) - f(t_0)}{t - t_0} $$ Then this limit is defined as the instantaneous velocity at time t0.

The Tangent Line Problem

In classical geometry, the tangent line to a circle is defined as "a straight line that intersects a circle at only one point". However, for more general curves, the definition "a straight line that intersects a curve at only one point" may not always apply.

For example, consider the curve $ y = x^3 $. At the origin, any line passing through the origin intersects the curve at only one point. However, only the x-axis is considered the true tangent at that point.

Let us now define the tangent line more generally. Let a curve C pass through a point M, and let N be another point on C. Draw a secant line MN. As point N moves toward M, the secant line rotates. If the limiting position of the secant line MN exists, we call that limiting line the tangent to curve C at point M.

Diagram showing secant and tangent lines to a curve

Now suppose curve C is represented by a function y = f(x), and M(x₀, y₀) is a point on this curve. Let y₀ = f(x₀). Take another point N(x, y) on the curve where y = f(x), then the slope of secant line MN is:

$$k = \frac{y - y_0}{x - x_0} = \frac{f(x) - f(x_0)}{x - x_0}$$ $$k = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}$$

This limit gives us the average rate of change or slope of secant line over an interval. If the time interval is short enough, this ratio approximates the instantaneous rate of change.

Let us now generalize this concept. Let a function y = f(x) be defined in a neighborhood of a point x₀. If:

$$\lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$$

exists, we call it the derivative of f(x) at x₀, and denote it as:

$$f'(x_0)$$

Also written as:

$$f'(x_0) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} \quad \text{or} \quad \left.\frac{dy}{dx}\right|_{x = x_0}$$

Alternatively, you may see:

$$f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}$$

or:

$$f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}$$

These are all equivalent definitions of the derivative.

Geometric Meaning of the Derivative

From the previous discussion on tangent lines and the definition of the derivative, we can conclude:

Let y = f(x) be a curve, and let M(x₀, f(x₀)) be a point on the curve. The derivative at point x₀, i.e., $ f'(x₀) $, is the slope of the tangent line at that point.

where α is the angle between the tangent and the x-axis.

If the derivative does not exist at that point, the curve has a vertical tangent line.
If $ f'(x₀) = 0 $, the tangent is horizontal, i.e., parallel to the x-axis.

Basic Differentiation Rules and Derivative Formulas

The basic derivative formulas of elementary functions and the differentiation rules discussed in this section are crucial in computing derivatives. We must practice and master them thoroughly. For easy reference, they are summarized below:

Derivatives of Constants and Basic Elementary Functions

Differentiation Rules for Sum, Difference, Product, and Quotient

Let u=u(x), v=v(x), and both functions are differentiable. Then:

Differentiation Rules for Sum, Difference, Product, and Quotient

Let y=f(u), u=g(x), and both f(u) and g(x) are differentiable. Then the composite function y=f(g(x)) has a derivative:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \quad \text{or} \quad y'(x) = f'(u) \cdot g'(x)$$

2. Introduction to Integrals

The Concept of Antiderivatives and Indefinite Integrals

Definition 1： If on an interval I, the function F(x) is differentiable and its derivative is f(x), that is, for every x∈ I, we have:

$$F'(x) = f(x) \quad \text{or} \quad dF(x) = f(x)dx$$ then F(x) is called an antiderivative of f(x) on interval I. For example, since sin'⁡(x)=cos(⁡x), it follows that sin(⁡x) is an antiderivative of cos⁡(x).

Definition 2: On interval I, the set of all antiderivatives of a function f(x), differing by arbitrary constants, is called the indefinite integral of f(x) on I, denoted as:

$$ \int f(x)\, dx$$

The symbol ∫ is called the integral sign.
The function f(x) is called the integrand.
The expression f(x)dx is called the integrand expression.
The variable x is called the variable of integration.

From the definition and previous explanation, if F(x) is an antiderivative of f(x) on interval I, then F(x) + C (where C is a constant) is an indefinite integral of f(x). That is:

$$\int f(x)\, dx = F(x) + C$$

Table of Basic Integrals

Since integration is the inverse operation of differentiation, it is natural that we can derive corresponding integral formulas from derivative formulas. Below is a list of several basic integral formulas, commonly referred to as the Table of Basic Integrals:

Examples of Definite Integral Problems

a. Area of a Curved-Edge Trapezoid

Let y = f(x) be a non-negative and continuous function on the interval [a, b]. The region bounded by the vertical lines x = a, x = b, the x-axis y = 0, and the curve y = f(x), is called a curved-edge trapezoid, with the curve as the upper boundary.

We know that for a rectangle with constant height, its area is calculated by the formula:

Area = height × width

However, for the curved-edge trapezoid, the height f(x) varies at each point on [a, b], so this area cannot be computed directly using the rectangle area formula. But since f(x) is continuous on [a, b], the change in height over a small subinterval is very small and can be approximated as constant.

So, if we divide [a, b] into many small intervals and approximate the height over each small interval using the function value at some point, we can estimate the area using a sum of small rectangles (each with width equal to the subinterval and height f(ξ)).

If we divide [a, b] into n subintervals:

$$a = x_0 < x_1 < \cdots < x_n = b$$

Then the subintervals are:

$$[x_0, x_1], [x_1, x_2], \dots, [x_{n-1}, x_n]$$

Each subinterval has width:

$$\Delta x_i = x_i - x_{i-1}, \quad i = 1, 2, \dots, n$$

For each subinterval [x_i−1, x_i], choose a point ξ_i ∈ [x_i−1, x_i], and approximate the sub-area with a rectangle of height f(ξ_i) and width Δx_i.

The total area A is approximately:

$$A \approx \sum_{i=1}^n f(\xi_i) \Delta x_i$$

To ensure all subinterval widths become arbitrarily small, let the maximum subinterval width tend to zero:

$$\lambda = \max\{\Delta x_1, \Delta x_2, \dots, \Delta x_n\} \to 0$$

Then, as n → ∞, the limit of the sum becomes the area under the curve:

$$A = \lim_{\lambda \to 0} \sum_{i=1}^n f(\xi_i) \Delta x_i$$

Definition of Definite Integral

Let the function f(x) be bounded on the interval [a, b]. Insert an arbitrary finite number of division points in the interval:

$$a = x_0 < x_1 < x_2 < \cdots < x_n = b$$

Divide the interval [a, b] into n subintervals:

$$[x_0, x_1], [x_1, x_2], \dots, [x_{n-1}, x_n]$$

Each subinterval has width:

$$\Delta x_1 = x_1 - x_0,\quad \Delta x_2 = x_2 - x_1,\quad \dots,\quad \Delta x_n = x_n - x_{n-1}$$

On each subinterval [x_i−1, x_i], pick any point ξ_i such that x_i−1 ≤ ξ_i ≤ x_i, and form the product f(ξ_i) · Δx_i for i = 1, 2, ..., n, then sum them to get:

$$S = \sum_{i=1}^n f(\xi_i) \Delta x_i$$

Let $ \lambda = \max\{\Delta x_1, \Delta x_2, \dots, \Delta x_n\} $. If the limit of this sum exists as $ \lambda \to 0 $, and is independent of how the partition $\{x_i\}$ and sample points $\{\xi_i\}$ are chosen, then we call this limit the definite integral of f(x) on the interval [a, b], denoted by:

$$\int_a^b f(x)\, dx = \lim_{\lambda \to 0} \sum_{i=1}^n f(\xi_i) \Delta x_i$$

Notation:

f(x): integrand (function to be integrated)
f(x) dx: integrand expression
x: integration variable
a: lower limit of integration
b: upper limit of integration
[a, b]: integration interval

Newton–Leibniz Formula

Now we introduce the Fundamental Theorem of Calculus, which provides a formula for calculating definite integrals using antiderivatives.

If the function F(x) is an antiderivative of the continuous function f(x) on the interval [a, b], then:

$$\int_a^b f(x)\, dx = F(b) - F(a)$$

6. Practice Problems

Test your understanding with these exercises. Solutions are hidden by default – try solving them first!

Problem 1

Given: $ y = 2x^3 - 5x^2 + 3x - 7 $

Find: $ y' $

Solution:

\[ \begin{aligned} y' &= (2x^3 - 5x^2 + 3x - 7)' \\ &= (2x^3)' - (5x^2)' + (3x)' - (7)' \\ &= 2 \cdot 3x^2 - 5 \cdot 2x + 3 - 0 \\ &= 6x^2 - 10x + 3 \end{aligned} \]

Problem 2

Given: $ f(x) = x^3 + 4 \cos x - \sin \frac{\pi}{2} $

Find: $ f'(x) $ and $ f'\left(\frac{\pi}{2}\right) $

Solution:

\[ f'(x) = 3x^2 - 4 \sin x \] \[ f'\left(\frac{\pi}{2}\right) = 3 \cdot \left(\frac{\pi}{2}\right)^2 - 4 = \frac{3}{4} \pi^2 - 4 \]

Problem 3

Given: $ y = e^x (\sin x + \cos x) $

Find: $ y' $

Solution:

\[ \begin{aligned} y' &= (e^x)'(\sin x + \cos x) + e^x(\sin x + \cos x)' \\ &= e^x (\sin x + \cos x) + e^x (\cos x - \sin x) \\ &= 2e^x \cos x \end{aligned} \]

Problem 4

Given: $ y = e^{x^3} $, find $ \frac{dy}{dx} $.

Solution:

We treat this as a composite function:

Let $ y = e^u $, where $ u = x^3 $

By the chain rule:

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = e^u \cdot 3x^2 = 3x^2 e^{x^3} \]

Problem 5

Given: $ y = \sin\left(\frac{2x}{1 + x^2}\right) $, find $ \frac{dy}{dx} $

Solution:

This is a composite function.

Let $ y = \sin u $, where $ u = \frac{2x}{1 + x^2} $

Then:

\[ \frac{dy}{du} = \cos u \quad \text{and} \quad \frac{du}{dx} = \frac{2(1 + x^2) - (2x)^2}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2} \] \[ \frac{dy}{dx} = \cos u \cdot \frac{2(1 - x^2)}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2} \cdot \cos\left(\frac{2x}{1 + x^2}\right) \]

Problem 6

Let $ y = \ln(\sin x) $, find $ \frac{dy}{dx} $.

Solution:

\[ \frac{dy}{dx} = (\ln(\sin x))' = \frac{1}{\sin x} \cdot (\sin x)' = \frac{\cos x}{\sin x} = \cot x \]

Problem 7

Let $ y = \sqrt[3]{1 - 2x^2} $, find $ \frac{dy}{dx} $.

Solution:

\frac{dy}{dx} = \left[(1 - 2x^2)^{\frac{1}{3}}\right]' = \frac{1}{3}(1 - 2x^2)^{-\frac{2}{3}} \cdot (-4x) = \frac{-4x}{3(1 - 2x^2)^{\frac{2}{3}}}

Problem 8

Let $ y = \ln(\cos(e^x)) $, find $ \frac{dy}{dx} $.

Solution:

Decompose the function as:
$ y = \ln u, \quad u = \cos v, \quad v = e^x $

Then:
$ \frac{dy}{du} = \frac{1}{u}, \quad \frac{du}{dv} = -\sin v, \quad \frac{dv}{dx} = e^x $

$ \frac{dy}{dx} = \frac{1}{u} \cdot (-\sin v) \cdot e^x = \frac{-\sin(e^x)}{\cos(e^x)} \cdot e^x = -e^x \tan(e^x) $

Or, directly:

\[ \frac{dy}{dx} = \left[\ln(\cos(e^x))\right]' = \frac{1}{\cos(e^x)} \cdot \left[\cos(e^x)\right]' = \frac{-\sin(e^x)}{\cos(e^x)} \cdot e^x = -e^x \tan(e^x) \]

Problem 9

Let $ y = e^{\sin\left(\frac{1}{x}\right)} $, find $ y' $.

Solution:

\[ y' = \left(e^{\sin\left(\frac{1}{x}\right)}\right)' = e^{\sin\left(\frac{1}{x}\right)} \cdot \cos\left(\frac{1}{x}\right) \cdot \left(\frac{1}{x}\right)' = e^{\sin\left(\frac{1}{x}\right)} \cdot \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \] \[ = -\frac{1}{x^2} \cdot e^{\sin\left(\frac{1}{x}\right)} \cdot \cos\left(\frac{1}{x}\right) \]

Problem 10

Let $y = \sin(nx)\;\cdot\;\sin^n x$ (where $n$ is a constant). Find $y'$.

Solution:

First, apply the product rule of differentiation:

\[ y' \;=\; (\sin(nx))'\,\cdot\,\sin^n x \;+\;\sin(nx)\,\cdot\,(\sin^n x)'. \]

To compute $(\sin(nx))'$ and $(\sin^n x)'$, we must use the chain rule for composite functions. Thus, we get:

\[ y' = n\cos(nx)\,\cdot\,\sin^n x \;+\; \sin(nx)\,\cdot\,n\sin^{n-1}x\,\cdot\,\cos x \]

\[ = n\sin^{n-1} x\;\bigl(\cos(nx)\,\sin x \;+\;\sin(nx)\,\cos x\bigr) \]

\[ = n\sin^{n-1} x\;\cdot\;\sin\bigl((n+1)x\bigr). \]

Problem 11

$y = x^3 + \dfrac{7}{x^4} - \dfrac{2}{x} + 12$. Find $y'$.

Solution:

y' = (x^3)' + (7x^{-4})' - (2x^{-1})' + (12)'\) = \(3x^2 - 28x^{-5} + 2x^{-2}\) = \(3x^2 - \dfrac{28}{x^5} + \dfrac{2}{x^2}

Problem 12

$y = 5x^3 - 2^x + 3e^x$. Find $y'$.

Solution:

y' = (5x^3)' - (2^x)' + (3e^x)'\) = \(15x^2 - 2^x\ln2 + 3e^x

Problem 13

$y = 2\tan x + \sec x - 1$. Find $y'$.

Solution:

y' = 2(\tan x)' + (\sec x)' - (1)'\) = \(2\sec^2 x + \sec x\tan x\) = \(2\cdot\dfrac{1+\sin x}{\cos^2 x}

Problem 14

$y = \sin x \cdot \cos x$. Find $y'$.

Solution:

y' = (\sin x)' \cos x + \sin x\,(\cos x)'\) = \(\cos x\cos x - \sin x\sin x\) = \(\cos^2 x - \sin^2 x\) = \(\cos 2x

Problem 15

$y = x^2\ln x$. Find $y'$.

Solution:

y' = (x^2)'\ln x + x^2\,( \ln x )'\) = \(2x\ln x + x^2\cdot\dfrac1x\) = \(2x\ln x + x

Problem 16

$y = 3e^x\cos x$. Find $y'$.

Solution:

y' = (3e^x)'\cos x + 3e^x\,(\cos x)'\) = \(3e^x\cos x - 3e^x\sin x

Problem 17

$y = \dfrac{\ln x}{x}$. Find $y'$.

Solution:

y' = \dfrac{(\ln x)'\cdot x - \ln x \cdot (x)'}{x^2}\) = \(\dfrac{1 - \ln x}{x^2}

Problem 18

$y = \dfrac{e^x}{x^2} + \ln3$. Find $y'$.

Solution:

y' = \Bigl(\dfrac{e^x}{x^2}\Bigr)' + (\ln3)'\) = \(\dfrac{e^x\cdot x^2 - e^x\cdot2x}{x^4}\) = \(\dfrac{e^x(x - 2)}{x^3}

Problem 19

$y = (2x + 5)^4$. Find $y'$.

Solution:

Let \(u = 2x + 5\). Then by the chain rule \[ \frac{dy}{dx} = \frac{dy}{du}\,\frac{du}{dx} = (u^4)' \cdot (2x + 5)' = 4u^3 \cdot 2 = 8(2x + 5)^3. \]

Problem 20

$y = \cos(4 - 3x)$.Find $y'$.

Solution:

Let \(u = 4 - 3x\). Then \[ \frac{dy}{dx} = \frac{dy}{du}\,\frac{du}{dx} = (\cos u)' \cdot (4 - 3x)' = -\sin u \cdot (-3) = 3\sin(4 - 3x). \]

Problem 21

Find:

\[ \int \sqrt{x}\,(x^2 - 5)\,dx \]

Solution:

\int \sqrt{x}\,(x^2 - 5)\,dx = \int \bigl(x^{\tfrac12}(x^2 -5)\bigr)\,dx = \int \bigl(x^{\tfrac52} -5x^{\tfrac12}\bigr)\,dx \] \[ = \frac{2}{7}\,x^{\tfrac{7}{2}} -5\cdot\frac{2}{3}\,x^{\tfrac{3}{2}} + C = \frac{2}{7}x^3\sqrt{x} -\frac{10}{3}x\sqrt{x} + C.

Problem 22

Find:

\[ \int \Bigl(\frac{x-1}{x^2}\Bigr)^3 dx \]

Solution:

\int \Bigl(\frac{x-1}{x^2}\Bigr)^3 dx = \int \frac{x^3 -3x^2 +3x -1}{x^2}\,dx = \int \bigl(x -3 + \tfrac{3}{x} - \tfrac{1}{x^2}\bigr)\,dx \] \[ = \int x\,dx -3\int dx +3\int \frac{dx}{x} -\int x^{-2}dx = \frac{x^2}{2} -3x +3\ln\lvert x\rvert + \frac{1}{x} + C.

Problem 23

Find:

\[ \int \bigl(e^x -3\cos x\bigr)\,dx \]

Solution:

\int (e^x -3\cos x)\,dx = \int e^x\,dx -3\int \cos x\,dx = e^x -3\sin x + C.

Problem 24

Calculate the area under $y=\sin x$ from $0$ to $\pi$:

Solution:

A = \int_{0}^{\pi}\sin x\,dx = \bigl[-\cos x\bigr]_{0}^{\pi} = \bigl(-\cos\pi\bigr)-\bigl(-\cos0\bigr) = (-(-1))-(-(1)) = 2.

Problem 25

Evaluate:

\[ \int_{0}^{4} \frac{x+2}{\sqrt{2x+1}}\,dx \]

Solution:

Let \(t = \sqrt{2x+1}\), then \(x=\tfrac{t^2-1}{2}\) and \(dx = t\,dt\). When \(x=0\), \(t=1\); when \(x=4\), \(t=3\). \[ \int_{0}^{4} \frac{x+2}{\sqrt{2x+1}}\,dx = \int_{1}^{3} \frac{\tfrac{t^2-1}{2}+2}{t}\,t\,dt = \tfrac12\int_{1}^{3}(t^2+3)\,dt \] \[ = \frac12\Bigl[\tfrac{t^3}{3}+3t\Bigr]_{1}^{3} = \tfrac12\bigl[(9+9)-( \tfrac13+3 )\bigr] = \tfrac12\bigl(18-\tfrac{10}{3}\bigr) = \frac{22}{3}. \]

Problem 26

$\displaystyle \int_{0}^{a} \bigl(3x^2 - x + 1\bigr)\,dx$

Solution:

\int_{0}^{a} (3x^2 - x + 1)\,dx = \Bigl[x^3 - \tfrac12 x^2 + x\Bigr]_{0}^{a} = a^3 - \tfrac12 a^2 + a = a\bigl(a^2 - \tfrac12 a + 1\bigr).

Problem 27

$\displaystyle \int_{1}^{2} \Bigl(x^2 + \tfrac1{x^4}\Bigr)\,dx$

Solution:

\int_{1}^{2} \bigl(x^2 + x^{-4}\bigr)\,dx = \Bigl[\tfrac13 x^3 - \tfrac13 x^{-3}\Bigr]_{1}^{2} = \bigl(\tfrac{8}{3} - \tfrac{1}{24}\bigr) - \bigl(\tfrac{1}{3} - \tfrac{1}{3}\bigr) = \tfrac{21}{8}.

Problem 28

$\displaystyle \int_{4}^{9} \sqrt{x}\,\bigl(1 + \sqrt{x}\bigr)\,dx$

Solution:

\int_{4}^{9} \bigl(x^{\tfrac12} + x\bigr)\,dx = \Bigl[\tfrac23 x^{\tfrac32} + \tfrac12 x^2\Bigr]_{4}^{9} = \bigl(18 + 40.5\bigr) - \bigl(\tfrac{16}{3} + 8\bigr) = \tfrac{271}{6}.

Problem 29

$\displaystyle \int_{\sqrt{3}}^{5} \frac{dx}{\,1 + x^2\,}$

Solution:

\int_{\sqrt{3}}^{5} \frac{dx}{1+x^2} = \bigl[\arctan x\bigr]_{\sqrt{3}}^{5} = \arctan(5) - \tfrac{\pi}{3}.

Problem 30

$\displaystyle \int_{-\tfrac12}^{\tfrac12} \frac{dx}{\sqrt{\,1 - x^2\,}}$

Solution:

\int_{-1/2}^{1/2} \frac{dx}{\sqrt{1 - x^2}} = \bigl[\arcsin x\bigr]_{-1/2}^{1/2} = \frac{\pi}{3}.

Problem 31: Multiply Two Complex Numbers

$\displaystyle \int_{0}^{\sqrt{3}\,a} \frac{dx}{\,a^2 + x^2\,}$

Solution:

\int_{0}^{\sqrt{3}a} \frac{dx}{a^2 + x^2} = \Bigl[\tfrac{1}{a}\arctan\!\bigl(\tfrac{x}{a}\bigr)\Bigr]_{0}^{\sqrt{3}a} = \frac{\pi}{3a}.

Problem 32

$\displaystyle \int_{0}^{1} \frac{dx}{\sqrt{\,4 - x^2\,}}$

Solution:

\int_{0}^{1} \frac{dx}{\sqrt{4 - x^2}} = \bigl[\arcsin(\tfrac{x}{2})\bigr]_{0}^{1} = \frac{\pi}{6}.

Problem 33

$\displaystyle \int_{-1}^{0} \frac{3x^4 + 3x^2 + 1}{x^2 + 1}\,dx$

Solution:

\int_{-1}^{0}\Bigl(3x^2 + \tfrac{1}{x^2+1}\Bigr)\,dx = \Bigl[x^3 + \arctan x\Bigr]_{-1}^{0} = 1 + \tfrac{\pi}{4}.

Problem 34

$\displaystyle \int_{-e^{-1}}^{-2} \frac{dx}{1 + x}$

Solution:

\int_{-e^{-1}}^{-2} \frac{dx}{1+x} = \bigl[\ln\lvert1+x\rvert\bigr]_{-e^{-1}}^{-2} = -1.

Problem 35

$\displaystyle \int_{0}^{\pi/4} \tan^2\theta\,d\theta$

Solution:

\int_{0}^{\pi/4}\tan^2\theta\,d\theta = \int_{0}^{\pi/4}(\sec^2\theta -1)\,d\theta = \bigl[\tan\theta - \theta\bigr]_{0}^{\pi/4} = 1 - \tfrac{\pi}{4}.

Problem 36

$\displaystyle \int_{0}^{2\pi} \lvert \sin x\rvert \,dx$

Solution:

\int_{0}^{2\pi}\lvert\sin x\rvert\,dx = \int_{0}^{\pi}\sin x\,dx + \int_{\pi}^{2\pi}(-\sin x)\,dx = 4.

Problem 37

$\displaystyle \int_{0}^{2} f(x)\,dx,$ where $f(x)= \begin{cases} x+1, & x\le1,\\ \tfrac12\,x^2, & x>1. \end{cases}$

Solution:

\int_{0}^{2} f(x)\,dx = \int_{0}^{1}(x+1)\,dx + \int_{1}^{2}\tfrac12\,x^2\,dx = \Bigl[\tfrac{x^2}{2}+x\Bigr]_{0}^{1} + \Bigl[\tfrac{x^3}{6}\Bigr]_{1}^{2} = \tfrac32 + \tfrac{7}{6} = \tfrac{8}{3}.