Notes
FTC Part 2
- If $F' = f$, then $\displaystyle \int_a^b f(x)\,dx = F(b) - F(a)$. Use an antiderivative without $+ C$.
- Examples: $\displaystyle \int_1^3 2x\,dx = 8$, $\displaystyle \int_0^\pi \sin x\,dx = 2$, $\displaystyle \int_1^e \frac{1}{x}\,dx = 1$.
FTC Part 1
- If $\displaystyle G(x) = \int_a^x f(t)\,dt$, then $G'(x) = f(x)$.
- With an upper limit $u(x)$: $\displaystyle \frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x))\,u'(x)$. Swapping limits adds a negative sign.
u-substitution
- Choose $u$ as an inside function whose derivative is also present. Rewrite in $u$, integrate, replace $u$.
- For definite integrals, convert the limits: use $u(a)$ and $u(b)$.
- Always differentiate your answer to check.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle \left[\frac{x^4}{4} + x\right]_0^2 = 4 + 2 = 6$.
FTC Part 1 gives the integrand evaluated at $x$.
With $u = x^2$, $du = 2x\,dx$, so $\displaystyle \int \cos u\,du = \sin u$.
$[\sin x]_0^{\pi/2} = 1 - 0 = 1$.
$\displaystyle \left[-\frac{1}{x}\right]_1^2 = -\frac{1}{2} + 1 = \frac{1}{2}$.
With $u = 3x$, $\displaystyle \int_0^3 \frac{e^u}{3}\,du = \frac{e^3 - 1}{3}$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
Let $F(x) = \int_0^x (t^2 - 1)\,dt$.
(a) Find $F'(x)$ and evaluate $F'(3)$.
Show the answer and scoring guide
By FTC Part 1, $F'(x) = x^2 - 1$. So $F'(3) = 8$.
Check off the points you earned:
(b) Find $F(3)$.
Show the answer and scoring guide
$\displaystyle F(3) = \left[\frac{t^3}{3} - t\right]_0^3 = (9 - 3) - 0 = 6$.
Check off the points you earned:
(c) Evaluate $\int_0^1 2x(x^2 + 3)^2\,dx$ using substitution.
Show the answer and scoring guide
Let $u = x^2 + 3$, $du = 2x\,dx$. The limits become 3 to 4. So $\displaystyle \int_3^4 u^2\,du = \left[\frac{u^3}{3}\right]_3^4 = \frac{64 - 27}{3} = \frac{37}{3}$.
Check off the points you earned: