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# Week 13 Problem Set A.
## Reading.
Chapter 5.4, 5.5, 5.6
On the Fundamental theorem of calculus (FTC I Existence theorem and FTC II Evaluation theorem), substitution method.
## Problems.
### Derivatives of integrals.
Use FTC I (Existence theorem) along with possibly other definite integral rules and chain rule to compute the following:
1. $\displaystyle \frac{d}{dx}\int_{0}^{x} \cos(t) \, dt$
2. $\displaystyle \frac{d}{dx}\int_{0}^{\sqrt{x}} \cos(t) \,dt$
3. $\displaystyle \frac{d}{dt}\int_{0}^{t^{4}} \sqrt{u}\,du$
4. $\displaystyle \frac{d}{d\theta} \int_{1}^{\sin(\theta)} 3 t^{2} \, dt$
Compute $dy / dx$ for each of the following. Again use FTC I to help you, and beware of chain rule and other derivative/integral rules that may be useful:
1. $\displaystyle y = \int_{1}^{x} \frac{1}{t}dt$
2. $\displaystyle y = \int_{\sqrt{x}}^{0}\sin(t^{2}) \, dt$
3. $\displaystyle y = x \int_{2}^{x^{2}} \sin(t^{3})\,dt$
4. $\displaystyle y = \int_{-1}^{x} \frac{t^{2}}{t^{2}+4} dt - \int_{3}^{x} \frac{t^{2}}{t^{2}+4} dt$
5. $\displaystyle y = \left(\int_{0}^{x} (t^{3}+1)^{10}dt\right)^{3}$
6. $\displaystyle y = \int_{\tan x}^{0} \frac{dt}{1+t^{2}}$
### Evaluating integrals.
Use a combination of FTC II (Evaluation theorem), geometry, and integration rules, if necessary, to evaluate the following integrals:
1. $\displaystyle \int_{-2}^{0} (2x+5)dx$
2. $\displaystyle \int_{0}^{2} x(x-3)dx$
3. $\displaystyle\int_{0}^{4}\left( 3x - \frac{x^{3}}{4} \right)dx$
4. $\displaystyle \int_{0}^{1} (x^{2} + \sqrt{x}) dx$
5. $\displaystyle \int_{0}^{\pi / 3} 2 \sec^{2}(x) dx$
6. $\displaystyle \int_{\pi / 4}^{3 \pi / 4} \csc \theta \cot \theta d\theta$
7. $\displaystyle \int_{\pi / 2}^{0} \frac{1+\cos(2t)}{2}dt$
8. $\displaystyle\int_{0}^{\pi / 4} \tan^{2}(x) dx$ .... Hint: Maybe trigonometric identities can help.
9. $\displaystyle \int_{0}^{\pi / 8} \sin(2x) dx$
10. $\displaystyle \int_{1}^{-1} (r+1)^{2}dr$
11. $\displaystyle \int_{\sqrt{2}}^{1}\left( \frac{u^{7}}{2}- \frac{1}{u^{5}} \right) du$
12. $\displaystyle \int_{1}^{\sqrt{2}} \frac{s^{2}+\sqrt{s}}{s^{2}} ds$
13. $\displaystyle \int_{\pi / 2}^{\pi} \frac{\sin(2x)}{2\sin(x)}dx$
14. $\displaystyle \int_{-4}^{4} |x| dx$
15. $\displaystyle\int_{1}^{8} \frac{(x^{1 / 3} + 1)(2 - x^{2 /3})}{x^{1 / 3}} dx$
16. $\displaystyle \int_{0}^{\pi / 3} (\cos x + \sec x)^{2}dx$
17. $\displaystyle \int_{0}^{\pi} \frac{1}{2}(\cos x + |\cos x|)dx$
### Areas.
With definite integrals, we can calculate the **signed areas** of a function, that is, the quantity $\int_{a}^{b}f(x)dx$ gives the signed area of the graph of $y=f(x)$ to the $x$-axis over the interval $[a,b]$, and if the region is above the $x$-axis, it is considered positive, while below the $x$-axis is negative.
However, sometimes we are interested in the **total geometric area**, in which case we want to treat the areas as only positive as usual.
Find the **total geometric area** of the shaded region in each of the following:
1. ![[1 teaching/smc-fall-2023-math-7/week-13/---files/Pasted image 20231124144607.png]]
2. ![[1 teaching/smc-fall-2023-math-7/week-13/---files/Pasted image 20231124144617.png]]
3. ![[1 teaching/smc-fall-2023-math-7/week-13/---files/Pasted image 20231124144953.png]]
4. ![[1 teaching/smc-fall-2023-math-7/week-13/---files/Pasted image 20231124145023.png]]
### Theory and practice.
1. Find a function $y(x)$ satisfying the following: $\displaystyle \frac{dy}{dx} = \sec(x)$ and $y(2)=3$. Express your answer $y(x)$ as an integral.
2. Find a function $y(x)$ satisfying the following: $\displaystyle \frac{dy}{dx} = \sqrt{1+x^{2}}$ and $y(1) = -2$. You may express your answer $y(x)$ as an integral.
3. Suppose that $\displaystyle \int_{1}^{x} f(t) dt = x^{2} -2x +1$. Find $f(x)$. Hint: Differentiate...
4. Suppose that $\displaystyle\int_{0}^{x} f(t) dt = x \cos(\pi x)$, find $f(4)$.
5. Find the linearization of $$
f(x) = 2 - \int_{2}^{x+1} \frac{9}{1+t}dt
$$ at $x=1$.
(By the way, the problems 1 and 2 here are called **initial value problems** in differential equations, which itself is a vast and wonderful subject.)
### (Optional) Another proof of FTC II Evaluation theorem.
In class we proved FTC II using FTC I and a corollary of MVT: That any two antiderivatives are equal up to an additive constant. There is a more direct way of proving it. Try it in the following three steps:
1. Let $a = x_{0} < x_{1} < x_{2} < \cdots < x_{n} =b$ be any partition of $[a,b]$, and let $F$ be any antiderivative of $f$. Show that $$
F(b) - F(a) = \sum_{i=1}^{n} F(x_{i}) - F(x_{i-1})
$$
2. Apply the Mean Value Theorem to each term to show that $F(x_{i}) - F(x_{i-1}) = f(c_{i})(x_{i}-x_{i-1})$ for some $c_{i}$ in the interval $(x_{i-1},x_{i})$. Then show that $F(b)-F(x)$ is a Riemann sum for $f$ on $[a,b]$.
3. From previous part and the definition of the definite integral, show that $$
F(b) - F(a) = \int_{a}^{b}f(x)dx
$$
Amazing!
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