Exercises

Evaluate $\int_1^3 x dx$.

---

$$\int_1^3 x dx = \frac{x^2}{2}\bigg\rvert_{x=1}^3 = \frac{3^2}{2} - \frac{1^2}{2} = 4$$

Evaluate $\int_1^3 \frac{x}{2} dx$.

---

$$\int_1^3 \frac{x}{2} dx = \frac{1}{2}\int_1^3 x dx = \frac{1}{2}\left(\frac{x^2}{2}\bigg\rvert_{x=1}^3\right) = \frac{1}{2}\left(\frac{3^2}{2} - \frac{1^2}{2}\right) = 2$$

Evaluate $\int_0^4 (x + 1) dx$.

---

$$\int_0^4 (x + 1) dx = \left(\frac{x^2}{2} + x\right)\bigg\rvert_{x=0}^4 = \left(\frac{4^2}{2} + 4\right) - \left(\frac{0^2}{2} + 0\right) = 12$$

Evaluate $\int_{1}^4 (x^3 - 1) dx$.

---

$$\int_{1}^4 (x^3 - 1) dx = \left(\frac{x^4}{4} - x\right)\bigg\rvert_{x=1}^4 = \left(\frac{4^4}{4} - 4\right) - \left(\frac{1^4}{4} - 1\right) = 60 - \left(-\frac{3}{4}\right) = \frac{243}{4}$$

Evaluate $\int_{-1}^1 x^{-4/5} dx$.

---

$$\int_{1}^3 x^{-4/5} dx = 5 x^{1/5} \bigg\rvert_{x=1}^3 = 5\cdot 3^{1/5} - 5\cdot 1^{1/5} = 5\cdot 3^{1/5} - 5 = 5(3^{1/5} - 1)$$

Evaluate $\int_1^4 \frac{1}{x} dx$.

---

$$\int_1^4 \frac{1}{x} dx = \ln x \bigg\rvert_{x=1}^4 = \ln 4 - \ln 1 = \ln 4$$

Evaluate $\int_1^{e^2} \frac{x+1}{x} dx$.

---

$$\int_1^{e^2}\frac{x+1}{x} dx = \int_1^{e^2} \left(1 + \frac{1}{x}\right) dx = \left(x + \ln x\right)\bigg\rvert_{x=1}^{e^2} = \left(e^2 + \ln e^2\right) - \left(1^2 + \ln 1\right) = e^2 + 2 - 1 = e^2 + 1 $$

Evaluate $\int_2^4 \frac{x+1}{x-1} dx$.

---

$$\int_2^4 \frac{x+1}{x-1}dx = \int_2^4 \frac{x}{x-1}dx + \int_2^4 \frac{1}{x-1} dx$$

Let's focus just on the first integral. Start by letting $u = x - 1$ so $du = dx$. This makes the integral $$\int_2^4 \frac{x+1}{x-1}dx = \int_2^4 \frac{x+1}{u}du$$

Now, we can get the x out of the numerator by substituting $x + 1= u + 2$.

$$\begin{aligned} \int_2^4 \frac{x+1}{x-1}dx &= \int_2^4 \frac{u+2}{u}du \\ &= \int_2^4 \left(\frac{u}{u} + \frac{2}{u}\right)du \\ &= \int_2^4 \left(1 + \frac{2}{u}\right)du \\ &= \left(u + 2\ln u\right)\bigg \rvert_{x=2}^4 \\ &= \left(x-1 + 2\ln (x-1)\right)\bigg \rvert_{x=2}^4 \\ &= 4 - 1 + 2\ln(4-1) - \left(2 - 1 + 2\ln(2 - 1)\right)\\ &= 3 + 2\ln(3) - 1 - 2\ln(1) \\ &= 2 + 2\ln 3 \end{aligned}$$

Evaluate $\int_{1}^4 (x + \sqrt{x}) dx$.

---

$$\begin{aligned} \int_1^{4} \left(x + \sqrt{x}\right) dx &= \int_1^{4} \left(x + x^{1/2}\right) dx \\ &= \left(\frac{x^2}{2} + \frac{2}{3}x^{3/2}\right)\bigg\rvert_{x=1}^{4} \\ &= \left(\frac{4^2}{2} + \frac{2}{3}\cdot4^{3/2}\right) - \left(\frac{1^2}{2} + \frac{2}{3}1^{3/2}\right) \\ &=\left(8 + \frac{16}{3}\right) - \left(\frac{1}{2} + \frac{2}{3}\right) \\ &= \frac{73}{6} \end{aligned}$$

Evaluate $\int_{1}^4 (\frac{3}{x} + \frac{x}{3}) dx$.

---

$$\int_{1}^4 \left(\frac{3}{x} + \frac{x}{3}\right) dx = \left(3\ln x + \frac{x^2}{6}\right) \bigg\rvert_{x=1}^4 = \left(3\ln 4 + \frac{4^2}{6}\right) - \left(3\ln 1 + \frac{1^2}{6}\right) = 3\ln 4 + \frac{16}{6} - \frac{1}{6} = 3 \ln 4 + \frac{5}{2}$$

Evaluate $\int_0^2 e^{x+1} dx$.

---

$$\int_0^2 e^{x+1} dx = e^{x+1}\bigg\rvert_{x=0}^2 = e^{2 + 1} - e^{0 + 1} = e^3 - e$$

Evaluate $\int_{-4}^1 2^{-2x} dx$.

---

First, we need to find an antiderivative using a substitution, $u = -2x$ so $du = -2dx$ or $dx = -\frac{1}{2}du$. That makes the integral

\begin{aligned} \int_{-4}^1 2^{-2x} dx &= \int_{-4}^1 2^u\left(-\frac{1}{2}du\right) \\ &= -\frac{1}{2} \int_{-4}^1 2^udu \\ &= -\frac{1}{2}\frac{2^u}{\ln 2}\bigg\rvert_{x=-4}^1 \\ &=-\frac{2^{-2x}}{2\ln 2}\bigg\rvert_{x=-4}^1 \\ &= -\frac{2^{-2}}{2\ln 2} - \left(-\frac{2^{8}}{2\ln 2}\right) \\ &= -\frac{1}{8\ln 2} + \frac{2^{10}}{8\ln 2} \\ &= \frac{2^{10} - 1}{8\ln 2} \end{aligned}

Evaluate $\int_0^{\pi} \sin x dx$.

---

$$\int_0^{\pi} \sin x dx = -\cos x\bigg\rvert_{x=0}^\pi = -\cos \pi -(-\cos 0) = -(-1) + 1 = 2$$

Evaluate $\int_{-\pi}^{\pi} \cos x dx$.

---

$$\int_{-\pi}^{\pi} \cos x dx = \sin x \bigg\rvert_{x=-\pi}^{\pi} = \sin \pi - \sin (-\pi) = 0$$

Evaluate $\int_{0}^{\pi/2} 2\sin (3x) dx$.

---

$$\int_{0}^{\pi/2} 2\sin (3x) dx = -\frac{2}{3}\cos (3x)\bigg\rvert_{x=0}^{\pi/2} = -\frac{2}{3}\left(\cos (3\pi/2) - \cos 0\right) = -\frac{2}{3}\left(0 - 1\right) = \frac{2}{3}$$

Evaluate $\int_{0}^{3} |x - 1| dx$.

---

Think of the function as

$$f(x) = \begin{cases} -(x+1), & x \lt 1 \\ x + 1, & x \ge 1 \end{cases}$$

Then split the integral into two parts, one from 0 to 1 and the other from 1 to 3.

Evaluate $\int_{0}^{3} |x - 1| dx$.

---

This function looks like $y = x - 1$ to the right of $x=1$ and $y=-(x - 1)$ to the left of $x=1$ so, to calculate the integral, I'll split it into two parts based on that division.

$$\begin{aligned} \int_{0}^{3} |x - 1| dx &= \int_{0}^{1} -(x-1) dx + \int_{1}^{3} (x-1) dx \\ &= \int_{0}^{1} (-x+1) dx + \int_{1}^{3} (x-1) dx \\ &= \left(-\frac{x^2}{2} + x\right)\bigg\rvert_{x=0}^1 + \left(\frac{x^2}{2} - x\right)\bigg\rvert_{x=1}^3 \\ &= \left(-\frac{1^2}{2} + 1\right) - \left(-\frac{0^2}{2} + 0\right) + \left(\frac{3^2}{2} - 3\right) - \left(\frac{1^2}{2} - 1\right) \\ &= \frac{1}{2} - 0 + \frac{3}{2} + \frac{1}{2} \\ &= \frac{5}{2} \end{aligned}$$

Evaluate $\int_{0}^{3} \sinh (3x) dx$.

---

$$\int_{0}^{3} \sinh (3x) dx = \frac{1}{3} \cosh(3x) \bigg\rvert_{x=1}^3 = \frac{1}{3} (\cosh 9 - \cosh 1)$$

Evaluate the expression $\frac{d}{dx}\int_{-2}^x e^{-t} dt$.

---

According to the Fundamental Theorem of Calculus this is just

$$\frac{d}{dx}\int_{-2}^x e^{-t} dt = e^{-x}$$

Evaluate the expression $\frac{d}{dx}\int_{2}^{x^2} (x + 1) dx$.

---

To see exactly what's happening here, let $F(x)$ be an antiderivative of $f(x) = x + 1$ then

$$\int_{2}^{x^2} (x + 1) dx = F(x^2) - F(2)$$

We can now get the derivative by using the Chain Rule.

$$\begin{aligned} \frac{d}{dx}\int_{2}^{x^2} (x + 1) dx &= \frac{d}{dx}(F(x^2) - F(2)) \\ &= \frac{d}{dx}F(x^2) - \frac{d}{dx}F(2) \\ &= \frac{d}{dx}F(x) \frac{d}{dx}(x^2) - 0 \\ &= f(x) \cdot 2x \\ &= (x + 1)(2x) \\ &= 2x^2 + 2x \end{aligned}$$

Evaluate the expression $\frac{d}{dx}\int_{-2}^{e^x} \frac{1}{x} dx$.

---

To see exactly what's happening here, let $F(x)$ be an antiderivative of $f(x) = \frac{1}{x}$ then

$$\int_{-2}^{e^x} \frac{1}{x} dx = F(e^x) - F(-2)$$

We can now get the derivative by using the Chain Rule.

$$\begin{aligned} \frac{d}{dx}\int_{-2}^{e^x} \frac{1}{x} dx &= \frac{d}{dx}(F(e^x) - F(-2)) \\ &= \frac{d}{dx}F(e^x) - \frac{d}{dx}F(-2) \\ &= \frac{d}{dx}F(x) \frac{d}{dx}(e^x) - 0 \\ &= f(x) \cdot e^x \\ &= \frac{1}{x} \cdot e^x \\ &= \frac{e^x}{x} \end{aligned}$$

Find the average value of the function $f(x) = 2\cos x$ on the interval $[0, \pi]$.

---

$$\begin{aligned} \text{average} &= \frac{1}{b - a}\int_{a}^b f(x) dx \\ &=\frac{1}{\pi - 0}\int_0^{\pi} 2\cos(x) dx \\ &=\frac{1}{\pi} \cdot 2\sin x \bigg\rvert_{x=0}^\pi \\ &= \frac{1}{\pi} (2\sin \pi - 2\sin 0) \\ &= 0 \end{aligned}$$

Find the average value of the function $f(x) = e^{-x}$ on the interval $[-1, 1]$.

---

$$\begin{aligned} \text{average} &= \frac{1}{b - a}\int_{a}^b f(x) dx \\ &=\frac{1}{1 - (-1)}\int_{-1}^1 e^{-x} dx \\ &=\frac{1}{2} (-e^{-x}) \bigg\rvert_{x=-1}^1 \\ &= \frac{1}{2}(-e^{-1} - (-e^{1})) \\ &= \frac{1}{2}(e^{1}-e^{-1}) \end{aligned}$$

Find the average value of the function $f(x) = x^2 + 1$ on the interval $[-2, 2]$.

---

$$\begin{aligned} \text{average} &= \frac{1}{b - a}\int_{a}^b f(x) dx \\ &=\frac{1}{2 - (-2)}\int_{-2}^2 \left(x^2 + 1\right) dx \\ &=\frac{1}{4} \left(\frac{x^3}{3} + x\right) \bigg\rvert_{x=-2}^2 \\ &= \frac{1}{4}\left( \left(\frac{2^3}{3} + 2\right) - \left(\frac{(-2)^3}{3} + (-2)\right) \right) \\ &= \frac{1}{4}\left( \frac{14}{3} - \left(-\frac{14}{3}\right) \right) \\ &= \frac{7}{3} \end{aligned}$$