Questions
- Calculate the area of each region \(R\) formed by bounding the following.
- \(y=x\), \(x=0\), \(x=6\) and the \(x\)-axis.
- \(y=-x+6\), \(x=0\), \(x=6\) and the \(x\)-axis.
- \(y=4\), \(x=-2\), \(x=3\) and the \(x\)-axis.
- \(y=\sqrt{36-x^2}\), \(x=-6\), \(x=6\) and the \(x\)-axis.
- \(y=3\sqrt{36-x^2}\), \(x=-6\), \(x=6\) and the \(x\)-axis.
- Approximate the region \(R\) given below by using \(n\) left-endpoint rectangles, \(n\) right-endpoint rectangles, and \(n\) midpoint rectangles.
- The region \(R\) formed by bounding \(y=x^3\) and \(x\)-axis on the interval \([0,1]\). Use \(n=6\) rectangles.
- The region \(R\) formed by bounding \(y=\frac{1}{x}\) and \(x\)-axis on the interval \([1,3]\). Use \(n=4\) rectangles.
- Calculate the area of each region \(R\) given below using the limit definition of area.
- The region \(R\) formed by bounding \(y=x^2\), the \(x\)-axis, \(x=1\) and \(x=2.\)
- The region \(R\) formed by bounding \(y=x^3\), the \(x\)-axis, \(x=0\) and \(x=1.\)
- The region \(R\) formed by bounding \(y=2x^2\), the \(x\)-axis, \(x=1\) and \(x=2.\)
- The region \(R\) formed by bounding \(y=-x^2+1\), the \(x\)-axis, \(x=-1\) and \(x=1.\)
- Calculate the following by interpreting the definite integral as signed area.
- \(\displaystyle\int_{1}^{2}x^2\,\mathrm{d}x\)
- \(\displaystyle\int_{0}^{1}x^3\,\mathrm{d}x\)
- \(\displaystyle\int_{1}^{2}2x^2\,\mathrm{d}x\)
- \(\displaystyle\int_{-1}^{1}(-x^2+1)\,\mathrm{d}x\)
- \(\displaystyle\int_{-6}^{6}\sqrt{36-x^2}\,\mathrm{d}x\)
- \(\displaystyle\int_{-6}^{6}3\sqrt{36-x^2}\,\mathrm{d}x\)
- Suppose that \(f\) and \(g\) are integrable and \[\displaystyle\int^{7}_{-1}f(x)\,\mathrm{d}x=3\quad\text{and}\quad \int^{7}_{-1}g(x)\,\mathrm{d}x=4.\] Calculate the following: \[\int_{-1}^{7}(-f(x)+3g(x)+4x)\,\mathrm{d}x.\]
- Suppose that \(f\) is integrable and \[3\leq f(x)\leq 5\] on the interval \([-1,2].\) Calculate an upper and lower bound for the following:
- \(\displaystyle\int_{0}^{1}f(x)\,\mathrm{d}x\)
- \(\displaystyle\int_{-1}^{2}f(x)\,\mathrm{d}x\)
- \(\displaystyle\int_{-1}^{2}5f(x)\,\mathrm{d}x\)
- \(\displaystyle\int_{-1}^{2}(f(x)+1)\,\mathrm{d}x\)
- \(\displaystyle\int_{-1}^{2}(f(x)+x)\,\mathrm{d}x\)
Answers
- \(18\)
- \(18\)
- \(20\)
- \(18\pi\)
- \(54\pi\)
- Right approximation is \(\frac{49}{144}\), left approximation is \(\frac{25}{144}\), midpoint approximation is \(\frac{71}{288}\)
- Right approximation is \(\frac{19}{20}\), left approximation is \(\frac{77}{60}\), midpoint approximation is \(\frac{3776}{3465}\)
- \(\frac{7}{3}\)
- \(\frac{1}{4}\)
- \(\frac{14}{3}\)
- \(\frac{4}{3}\)
- \(\frac{7}{3}\)
- \(\frac{1}{4}\)
- \(\frac{14}{3}\)
- \(\frac{4}{3}\)
- \(18\pi\)
- \(54\pi\)
- \(105\)
- \(3\leq\displaystyle\int_{0}^{1}f(x)\,\mathrm{d}x\leq 5\)
- \(9\leq \displaystyle\int_{-1}^{2}f(x)\,\mathrm{d}x\leq 15\)
- \(45\leq \displaystyle\int_{-1}^{2}5f(x)\,\mathrm{d}x\leq 75\)
- \(12\leq\displaystyle\int_{-1}^{2}(f(x)+1)\,\mathrm{d}x\leq 18\)
- \(10.5\leq \displaystyle\int_{-1}^{2}(f(x)+x)\,\mathrm{d}x \leq 16.5\)