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Practice Problems for 4/24

  1. Write the following limit as a definite integral
    \[ \lim_{||P||\rightarrow 0} \sum_{k=1}^n \sqrt{c_k}, \] where \(P\) is a partition of the interval [5,10].
    2. \[ \int_5^{10} \sqrt{x}\,dx \]
  2. In class we found the limit of the Riemann sums of \(f(x)=x^2\) on [0,1] with \(n\) equal subintervals using both left and right endpoints for the \(c_k\) is equal to 1/3. From this information, and from what you know about the graph of \(x^2\), determine the value of the following definite integrals.
    1. \( \int_0^1 x^2\, dx =\frac{1}{3}\), since the limit of a Riemann sum is equal to the definite integral, which we can interpret geometrically as the signed area under a curve.
    2. \( \int_{-1}^0 x^2\, dx=\frac{1}{3}\), by symmetry since \(x^2\) is an even function.
    3. \( \int_{-1}^1 x^2\, dx=\int_{-1}^0 x^2\, dx + \int_0^1 x^2\,dx=\frac{1}{3}+\frac{1}{3}=\frac{2}{3},\) where we have used parts (a) and (b) along with the additive property of integrals.

  3. Find the value of \(\int_{-2}^3 |x|\, dx\) using the geometric interpretation of definite integrals, and any relevant properties of definite integrals.
    4. Since \(|x|=x\) for \(x\geq 0\) and \(|x|=-x\) for \(x\leq 0\), we need to break the interval [-2,3] into two intervals at \(x=0\) in order to integrate the function. Again, using the additive property, we have \[ \int_{-2}^3|x|\,dx=\int_{-2}^0|x|\,dx+ \int_0^3|x|\,dx=\int_{-2}^0 -x\,dx+ \int_0^3 x\,dx. \] To find the value of each integral, find the area under the curve using the fact that each is a triangle. \[ \int_{-2}^0 -x\,dx=\frac{1}{2}(2)(2)\quad \textrm{and}\quad \int_0^2 x\,dx=\frac{1}{2}(3)(3). \] Therefore, \(\int_{-2}^3 |x|\, dx=\frac{13}{2}\).
  4. Suppose \(f\) is a function which is continuous for all \(x\). Given \(\int_0^{20}f(x)\, dx=15\) and \(\int_0^{12} f(x)\,dx=7\) find the value of \(\int_{12}^{20}\frac{1}{2}f(x)\, dx\).
    5. This will also use the additive property of integrals, but it may look at bit strange. Technically, \(\int_a^b f(x)\, dx=\int_a^cf(x)\,dx+\int_c^b f(x)\, dx\) for any \(c\). That is \(c\) is not necessarily between \(a\) and \(b\). We'll use this here with \(a=12\), \(b=20\) and \(c=0\). \[ \int_{12}^{20} f(x)\,dx=\int_{12}^0 f(x)\,dx +\int_0^{20}f(x)\,dx. \] To find the value of the first integral on the right side, we'll reverse the order of integration \[ \int_{12}^0 f(x)\,dx=-\int_0^{12} f(x)\,dx=-7. \] Therefore, using the constant multiple property, \(\int_{12}^{20} \frac{1}{2}f(x)\,dx=\frac{1}{2}\int_{12}^{20}f(x)\,dx=\frac{1}{2}(-7+15)=4.\)