5.2: The Definite Integral
- Page ID
- 144293
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Express \(\displaystyle \lim_{n \to \infty} \sum_{i = 1}^{n} (x_i^*) \Delta x\) over \([1, 3]\) as a definite integral.
- Express \(\displaystyle \lim_{n \to \infty} \sum_{i = 1}^{n} \sin^2(2\pi x_i^*) \Delta x\) over \([0, 1]\) as a definite integral.
- Evaluate \(\displaystyle \int_{-2}^4 5\ dx\) using formulas from geometry.
- Evaluate \(\displaystyle \int_1^3 (3 - x)\ dx\) using formulas from geometry.
- Evaluate \(\displaystyle \int_{-2}^1 (2x + 1)\ dx\) using formulas from geometry.
- Evaluate \(\displaystyle \int_{-1}^2 (2 - |x|)\ dx\) using formulas from geometry.
- Evaluate \(\displaystyle \int_{-3}^3 \sqrt{9 - x^2}\ dx\) using formulas from geometry.
- Given that \(\displaystyle \int_0^1 x\ dx = \dfrac{1}{2}\), \(\displaystyle \int_0^1 x^2\ dx = \dfrac{1}{3}\), and \(\displaystyle \int_0^1 x^3\ dx = \dfrac{1}{4}\), evaluate \(\displaystyle \int_0^1 (1 - 2x - 3x^2 + 8x^3)\ dx\).
- Given that \(\displaystyle \int_0^1 x\ dx = \dfrac{1}{2}\), \(\displaystyle \int_0^1 x^2\ dx = \dfrac{1}{3}\), and \(\displaystyle \int_0^1 x^3\ dx = \dfrac{1}{4}\), evaluate \(\displaystyle \int_0^1 (1 - 2x)^3\ dx\).
- Use a Riemann sum with the formulas \(\displaystyle \sum_{i=1}^n 1 = n\) and \(\displaystyle \sum_{i=1}^n i = \dfrac{n(n+1)}{2}\) to evaluate \(\displaystyle \int_0^4 (2x - 5)\ dx\).
- Use a Riemann sum with the formulas \(\displaystyle \sum_{i=1}^n i = \dfrac{n(n+1)}{2}\) and \(\displaystyle \sum_{i=1}^n i^2 = \dfrac{n(n+1)(2n+1)}{6}\) to evaluate \(\displaystyle \int_{-2}^{2} (3x^2 - 2x)\ dx\).