The Fundamental Theorem of Calculus

$\dfrac{x^2 - 3}{\sqrt{x}} = x^{3/2} - 3x^{-1/2}$.

$\dfrac{2}{5}x^{5/2} - 3 \cdot 2x^{1/2} = \dfrac{2}{5}x^{5/2} - 6\sqrt{x}$.

$\dfrac{2}{5}x^{5/2} - 6\sqrt{x} + C$.

Derivative: $x^{3/2} - 3x^{-1/2}$ ✓.

$\dfrac{1}{9 + 4x^2} = \dfrac{1}{4}\cdot\dfrac{1}{\frac{9}{4} + x^2}$, so $a = \tfrac{3}{2}$.

$\dfrac{1}{4}\cdot\dfrac{2}{3}\arctan\dfrac{2x}{3} = \dfrac{1}{6}\arctan\dfrac{2x}{3}$.

$\dfrac{1}{6}\arctan\dfrac{2x}{3} + C$.

Derivative: $\dfrac{1}{6}\cdot\dfrac{2/3}{1 + 4x^2/9} = \dfrac{1}{9 + 4x^2}$ ✓.

$4\sin^2 x = 2(1 - \cos 2x) = 2 - 2\cos 2x$.

$F(x) = 2x - \sin 2x + C$.

$F(0) = 0 - 0 + C = 1$, so $C = 1$.

$F'(x) = 2 - 2\cos 2x = 4\sin^2 x$ ✓.

A polynomial minus $\cos x$: continuous everywhere.

$F(x) = 2x^3 - \sin x$.

$F\left(\tfrac{\pi}{2}\right) - F(0) = 2\cdot\dfrac{\pi^3}{8} - 1 - 0 = \dfrac{\pi^3}{4} - 1$.

$\sqrt{x} \neq 0$ on $[1, 4]$, so the integrand is continuous there.

$\dfrac{x + 1}{\sqrt{x}} = x^{1/2} + x^{-1/2}$, so $F(x) = \dfrac{2}{3}x^{3/2} + 2x^{1/2}$.

$F(4) - F(1) = \left(\dfrac{16}{3} + 4\right) - \left(\dfrac{2}{3} + 2\right) = \dfrac{20}{3}$.

Both pieces are continuous on their parts, so split at $1$: $\displaystyle\int_0^1 x^2\,dx + \int_1^{e}\frac{dx}{x}$.

Antiderivatives $\dfrac{x^3}{3}$ and $\log x$.

$\left(\dfrac{1}{3} - 0\right) + (\log e - \log 1) = \dfrac{1}{3} + 1 = \dfrac{4}{3}$.

$\dfrac{1}{(x - 2)^2}$ is undefined at $x = 2$, which is inside $[0, 3]$. Not continuous.

Carrying on anyway: $-\dfrac{1}{x - 2}$ is an antiderivative only on pieces that avoid $x = 2$.

The formula would give $F(3) - F(0) = -1 - \tfrac{1}{2} = -\tfrac{3}{2}$: negative for a positive integrand, so it is wrong.

$f(t) = \sqrt{1 + t^3}$, $\ g(x) = 1$, $\ h(x) = x^2$.

$\sqrt{1 + (x^2)^3}\cdot 2x$.

The lower limit is constant: subtract $0$.

$F'(x) = 2x\sqrt{1 + x^6}$.

$f(t) = e^{t^2}$, $\ g(x) = \sin x$, $\ h(x) = 3$.

The upper limit is constant: $0$.

$-\,e^{\sin^2 x}\cdot\cos x$.

$G'(x) = -\cos x\,e^{\sin^2 x}$.

$f(t) = \log(1 + t^2)$, $\ g(x) = x$, $\ h(x) = x^3$.

$\log(1 + x^6)\cdot 3x^2$.

$-\log(1 + x^2)\cdot 1$.

$H'(1) = 3\log 2 - \log 2 = 2\log 2$.

Integrand $f(t)$, limits $0$ and $x$. Differentiate both sides of the equation.

Left side: $f(x)\cdot 1$. Right side: $(x^2e^{x})' = (2x + x^2)e^{x}$.

The lower limit $0$ is constant: $0$. So $f(x) = (2x + x^2)e^{x}$.

$f(1) = 3e$.

$F'(x) = \dfrac{9 - x^2}{1 + x^2}$.

$1 + x^2 > 0$, so $F'(x) = 0$ at $x = \pm 3$.

$F' < 0$ on $(-\infty, -3)$, $\ F' > 0$ on $(-3, 3)$, $\ F' < 0$ on $(3, \infty)$.

Increasing on $(-3, 3)$; decreasing on $(-\infty, -3)$ and $(3, \infty)$.

$G'(x) = \dfrac{1}{2 + \sin(x^2 - 6x)}\cdot(2x - 6)$.

$2 + \sin(\cdot) \ge 1 > 0$, so $G'(x) = 0$ only at $x = 3$.

$G' < 0$ for $x < 3$ and $G' > 0$ for $x > 3$.

It is the only sign change, from $-$ to $+$. So the absolute min is at $x = 3$, where $x^2 - 6x = -9$.

$\Delta x = \dfrac{2}{n}$, $\ x_j = 1 + \dfrac{2j}{n}$.

$\displaystyle\int_1^3 x^3\,dx$.

$\left[\dfrac{x^4}{4}\right]_1^3 = \dfrac{81 - 1}{4} = 20$.

$\Delta x = \dfrac{1}{n}$, $\ x_j = \dfrac{j}{n}$.

$\displaystyle\int_0^1 \frac{dx}{1 + x^2}$.

$\arctan 1 - \arctan 0 = \dfrac{\pi}{4}$.

$\dfrac{j}{n^2} = \dfrac{j}{n}\cdot\dfrac{1}{n}$: $\Delta x = \dfrac{1}{n}$, $\ x_j = \dfrac{j}{n}$, term $x_j e^{x_j^2}\,\Delta x$.

$\displaystyle\int_0^1 x e^{x^2}\,dx$.

$\dfrac{d}{dx}\tfrac{1}{2}e^{x^2} = xe^{x^2}$, so $\left[\tfrac{1}{2}e^{x^2}\right]_0^1 = \dfrac{e - 1}{2}$.