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math for fun week 1
This week, we look at problems 1-11 of the MIT integration bee.
Question 1
$$ \int \frac{x + \sqrt{x}}{1 + \sqrt{x}} dx $$
Rationalize the denominator:
$$ \frac{x+\sqrt{x}}{1+\sqrt{x}} \cdot \frac{1-\sqrt{x}}{1-\sqrt{x}} = \frac{\cancel{x} + \sqrt{x}-x\sqrt{x} \cancel{- x}}{1-x}=\frac{\cancel{(1-x)}\sqrt{x}}{\cancel{1-x}} = \sqrt{x} $$
Therefore, our integral is:
$$ \int \sqrt{x} dx = \boxed{\frac{2}{3}x^{3/2} + C} $$
Question 2
$$ \int \frac{e^{x+1}}{e^x + 1} dx $$
Let $u = e^x + 1$, so $du = e^x dx$. Also rewrite the numerator as $e^{x + 1} = e \cdot e^x$
$$ \int\frac{e\cdot e^x dx}{e^x + 1} = e\int\frac{du}{u}=e\ln|u| + C=\boxed{e\ln(e^x + 1) + C} $$
Note: we omit the absolute value in our answer because $e^x + 1 > 0$ for all $x \in \mathbb{R}$
Question 3
$$ \int \sqrt[3]{3\sin(x) - \sin(3x)}dx $$
Recall the triple angle identity:
$$ \sin(3\theta) = -4\sin^3\theta + 3\sin\theta $$
So we can rewrite the integrand as:
$$ \int\sqrt[3]{\cancel{3\sin(x)}+4\sin^3 x\cancel{-3\sin x}}dx=\int\sqrt[3]{4\sin^3 x} = \sqrt[3]4 \int \sin xdx = \boxed{-\sqrt[3]4 \cdot \cos x + C} $$
Question 4
$$ \int_1^{e^e} \frac{\ln(x^{\ln(x^x)})}{x^2}dx $$
Rewrite the numerator as $\ln(x^{\ln(x^x)}) = \ln(x^{x\ln x})$, then let $u = \ln x$, $du = \frac{1}{x} dx$, so our integrand becomes:
$$ \frac{\ln{x^{ux}}}{x}du = \frac{u\cancel{x}\ln x}{\cancel{x}}du = u\ln x du = u^2 du $$
Adjusting the bounds of integration as $\ln 1 = 0$ and $\ln e^e = e$, our solution is:
$$ \int_0^{e} u^2du = \frac{1}{3}[u^3]_0^e = \boxed{\frac{e^3}{3}} $$
Question 5
$$ \int_{-\pi/2}^{\pi/2} \cos(20x)\sin(25x)dx $$
Recall that cosine is even, so its values in $-x$ are equal to those in $+x$.
However, sine is odd so its values in $-x$ and $+x$ are inverses. So our integrand is odd.
Notice that since our bounds are $\pm \frac{\pi}{2}$, and the integrand is odd, then our integral is:
$$ \boxed{0} $$
Question 6
$$ \int_0^{2\pi} \sin x \cos x \tan x \cot x \sec x \csc x dx $$
Here's the fun part: everything cancels out.
$$ \int_0^{2\pi} \cancel{\sin x} \cancel{\cos x} \cancel{\frac{\sin x}{\cos x}} \cancel{\frac{\cos x}{\sin x}} \cancel{\frac{1}{\cos x}} \cancel{\frac{1}{\sin x}}dx = \int_0^{2\pi} dx = [x]_0^{2\pi} = \boxed{2\pi} $$
Question 7
$$ \int \frac{x\ln(x)\cos(x) - \sin(x)}{x\ln^2x}dx $$
The key here is recognizing the quotient rule.
$$ f(x)=\frac{\sin x}{\ln x}, f'(x) = \frac{\ln(x)\cos(x) - \sin(x) \cdot \frac{1}{x}}{\ln^2x} $$
Multiply both sides of the fraction by $x$:
$$ f'(x)=\frac{x\ln(x)\cos(x)-\sin(x)}{x\ln^2x} $$
Which matches our integrand, so by the Fundamental Theorem of Calculus, our answer is simply $f(x)$.
$$ \boxed{\frac{\sin x}{\ln x} + C} $$
Question 8
$$ \int_1^2 (2^{x-1} + \log_2(2x))dx $$
By change of base formula:
$$ \log_2(2x) = \log_2(2) + \log_2(x) = 1 + \frac{\ln x}{\ln 2} $$
So we can evaluate our integral:
$$ \int_1^2 (2^{x-1} + \frac{\ln x}{\ln 2} + 1)dx = [\frac{2^{x-1} + x\ln x - x}{\ln 2} + x]_1^2 = \frac{\cancel{2^1 - 2^0} + 2\cancel{\ln2} - \cancel{1\ln1} \cancel{- 2 + 1}}{\cancel{\ln 2}}+1 $$
And so our final answer is:
$$ 2 + 1 = \boxed{3} $$
Note that we can derive the integral of $\ln x$ via integration by parts:
$$ \int \ln x dx = x\ln x - \int dx = x \ln x - x + C $$
Question 9
$$ \int_0^1 x^{2024}(1-x^{2025})^{2025}dx $$
Let $u = 1 - x^{2025}$, $du = -2025x^{2024} dx$, so $x^{2024}dx = -\frac{du}{2025}$.
After adjusting the bounds of integration, the negatives cancel and the bounds are unchanged.
$$ \frac{1}{2025}\int_0^1u^{2025} du = \frac{1}{2025 \cdot 2026}[u^{2026}]_0^1 = \boxed{\frac{1}{4102650}} $$
Math for Fun! Evaluating 2025 x 2026 quickly
$$ n(n+1) = n^2 + n $$
If $n = 2025$, then our product is $2025^2 + 2025$.
How do we do $2025^2$ quickly? We know $2025 = 2000 + 25$, so:
$$ (2000 + 25)^2 = 2000^2 + 2\cdot 2000\cdot25 + 25^2 = 4,000,000 + 100,000 + 625 = 4,100,625 $$
Add 2025:
$$ 4,100,625 + 2025 = 4,102,650 $$
Question 10
$$ \int_0^{10} x(x-\frac{1}{2})(x-1)dx $$
Expanding the polynomial results in a trinomial, which is harder to integrate.
We can do better. Notice how the integrand is a cube centered around $x = \frac{1}{2}$
We can use that to our advantage.
Let $u = x - \frac{1}{2}$, $du = dx$, $x = u + \frac{1}{2}$, then expand:
$$ \int_{-1/2}^{19/2} u(u+\frac{1}{2})(u - \frac{1}{2})du = \int_{-1/2}^{19/2} u(u^2 - \frac{1}{4}) du = [\frac{u^4}{4}-\frac{u^2}{8}]_{-1/2}^{19/2} $$
Expanding our final result may seem hard:
$$ \frac{19^4}{64}-\frac{19^2}{32}-\frac{1}{64}+\frac{1}{32} $$
Convert to a common denominator and combine:
$$ \frac{19^4-2\cdot19^2+1}{64}=\frac{(19^2-1)^2}{64} = (\frac{(19+1)(19-1)}{8})^2 = (\frac{20\cdot18}{8})^2 = 45^2 = \boxed{2025} $$
Question 11
$$ \int_0^{20}\lceil \frac{\lfloor x \rfloor}{2} \rceil dx $$
If you've never done floor and ceiling in integrals before, think of them as summing rectangles.
At $x = 20$, the value of the function is $10$. So we are going to sum rectangles up to 10.
What is the length of each rectangle? It's going to be 2, because we divided by 2.
One more condition: at $x = 0$, our function is zero. Going left or right, it will floor down to either 0 or -1.
However, dividing -1 by 2 gives -0.5, which will ceiling back to zero.
So our rectangles start from 1 and go to 3, so on and so on until we get from 19 to 21.
Remembering that we only take half of the last rectangle:
$$ 2(1)+2(2)+2(3)+2(4)+2(5)+2(6)+2(7)+2(8)+2(9)+1(10) = 2(1 + 2+...+9) + 10 $$
Think about that middle sum.
$1 + 9 = 10$, $2 + 8 = 10$, and so on until $4 + 6 = 10$ and $5$ is left over. So that adds up to $45$.
$$ 2(45) + 10 = 90 + 10 = \boxed{100} $$