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Approximation of Integrals


  1. We know that we will set up subdivisions every $\frac{1}{2}$ units, since $\Delta x = \frac{2}{4}=\frac{1}{2}$. When estimating, you should always make a table. Doing so, we get the following: \begin{array}{c|c|c|c|c|c} x & 0 & \frac{1}{2} & 1 & \frac{3}{2} & 2 \\ \hline f(x) & 0 & \frac{3}{4} & 1 & \frac{3}{4} & 0 \end{array} since this is a right-hand Riemann sum, we know that the heights of the rectangles we use to estimate will be the values of $f(x)$ starting from $\frac{3}{4}$, as starting from zero would give a left-hand estimate. Now all we have to do is take the sums of the areas of the rectangles, where the $\Delta x$ is the base and $f(x)$ is the height. So, we sum $\frac{1}{2}\left(\frac{3}{4}+1+\frac{3}{4}+0\right)=\frac{10}{8}=\frac{5}{4}$.

  2. Since $n=6$, $\Delta x = \frac{2}{6}=\frac{1}{3}$. This means that we will be sampling every $f(x)$ that appears for every $\frac{1}{3}$ increase in $x$. Once again, if we make a table, we can simply plug the values into the Simpson's Rule formula. Doing so, we get \begin{array}{c|c|c|c|c|c|c|c} x & 0 & \frac{1}{3} & \frac{2}{3} & 1 & \frac{4}{3} & \frac{5}{3} & 2 \\ \hline f(x) & 0 & \frac{1}{27} & \frac{8}{27} & 1 & \frac{64}{27} & \frac{125}{27} & 2 \end{array} Now that we have our values, just use the formula! $\frac{1}{9}\left(0+\frac{4}{27}+\frac{16}{27}+4+\frac{128}{27}+\frac{500}{27}+8\right)=\frac{1}{9}\cdot\frac{972}{27}=\frac{1944}{486}=4$.

  3. This one's easy: the table is already given! From the table, we can see that $\Delta x=2$, so all we have to do is plug it into the trapezoidal rule formula: $\frac{1}{2}\cdot2(2+2\cdot3+2\cdot3+2\cdot4+5)=2+6+6+8+5=27$.

  4. As with all approximations, we first need to determine $\Delta x$. In this case, it'll be $\frac{2-0}{4}=\frac{1}{2}$. However we know that we won't be starting at zero or $\frac{1}{2}$, but exactly between the two, at $\frac{1}{4}$, as this is a midpoint Riemann sum. Then increasing $x$ by $\frac{1}{2}$, we can make the following table of values: \begin{array}{c|c|c|c|c} x & \frac{1}{4} & \frac{3}{4} & \frac{5}{4} & \frac{7}{4} \\ \hline f(x) & \frac{33}{16} & \frac{41}{16} & \frac{57}{16} & \frac{81}{16} \end{array} Now, we simply sum up the rectangles: $\frac{1}{2}\left(\frac{33}{16}+\frac{41}{16}+\frac{57}{16}+\frac{81}{16}\right)=\frac{212}{32}=\frac{53}{8}$.

Integrals with the FTC

  1. By using the first Fundamental Theorem, we know that $\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\sec^2xdx=\displaystyle\int\sec^2xdx\biggr\rvert_{\frac{\pi}{6}}^{\frac{\pi}{3}}$. So, this integral is equal to $\tan x\biggr\rvert_{\frac{\pi}{6}}^{\frac{\pi}{3}}=\tan \frac{\pi}{3}-\tan\frac{\pi}{6}=\sqrt{3}-\frac{1}{\sqrt{3}}=\frac{3\sqrt{3}}{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}$.

  2. Evaluating the indefinite integral using u-substitution, we can let $u=\sin x$ and $du=\cos x$. Hence, the integral becomes $\displaystyle\int_0^{\frac{\pi}{2}}\sin{u}du=-\cos{u}\biggr\rvert_0^{\frac{\pi}{2}}=-\cos(\sin(x))\biggr\rvert_0^{\frac{\pi}{2}}=-\cos(1)+1$.

  3. Looking at the function inside the absolute value bars, we can see that the zero of that function occurs at $x=2$. Also notice that the domain of the integral is from -3 to 5. Recall the definition of the absolute value:
    $|x|=$ $\begin{cases} x & \text{for $x>0$}\\ -x & \text{for $x< 0$} \end{cases}$
    Knowing this, we can rewrite our integrand as the following:
    $3-|x-2|=$ $\begin{cases} 3-(x-2) & \text{for $2\leq x\leq5$}\\ 3+(x-2) & \text{for $-3\leq x\leq2$} \end{cases}$
    Thus, we can rewrite our integral as the sum of two separate integrals: $\displaystyle\int_3^5(3-|x-2|)dx=$ $\displaystyle\int_{-3}^2(3+(x-2))dx+\displaystyle\int_2^5(3-(x-2))dx=$ $\displaystyle\int_{-3}^2(x+1)dx+\displaystyle\int_2^5(5-x)dx$. Now, we simply evaluate using the first Fundamental Theorem of Calculus: $\displaystyle\int_{-3}^2(x+1)dx+\displaystyle\int_2^5(5-x)dx=$ $\left(\frac{x^2}{2}+x\right)\biggr\rvert_{-3}^2+\left(5x-\frac{x^2}{2}\right)\biggr\rvert_2^5=$ $\left((2+2)-\left(\frac{9}{2}-3)\right)+\left((25-\frac{25}{2})\right)-(10-2)\right)$ $=7-\frac{9}{2}+17-\frac{25}{2}=$ $\frac{48}{2}-\frac{34}{2}=\frac{14}{2}=7$.

  4. Using u-substitution with $u=x-1$ and $du=1$ to evaluate the antiderivative, we get that this integral is equal to $4\displaystyle\int_2^{e+1}\frac{1}{u}du=4\ln|u|\biggr\rvert_2^{e+1}=4\ln|x-1|\biggr\rvert_2^{e+1}$. Evaluating using FTC, we get the answer: $(4\ln(e))-(4\ln(1))=4\ln(e)-0=4$.

  5. Using u-substitution with $u=\ln(\pi x)$ and $du=\frac{1}{x}$, we can rewrite this integral as $\displaystyle\int_{\frac{1}{\pi}}^{\frac{e^4}{\pi}}{u}du=\frac{u^2}{2}\biggr\rvert_{\frac{1}{\pi}}^{\frac{e^4}{\pi}}=\frac{\ln^2(\pi x)}{2}\biggr\rvert_{\frac{1}{\pi}}^{\frac{e^4}{\pi}}=\frac{\ln^2(e^4)}{2}-\frac{\ln1}{2}=\frac{16}{2}-0$ $=8$.

  6. Since any definite integral produces a numerical value, the derivative of a definite integral will always be 0. Hence, the first function evaluates to 0. The second function, once its interior is derived, becomes $\displaystyle\int_0^32dx=2x\biggr\rvert_0^3=6$. Thus, the positive difference between 0 and 6 is 6.

Fun Integrals

  1. The form of this problem should be a dead giveaway that it is a Riemann sum. When dealing with such, we first try to identify $\Delta x$. This term will be a fraction with $n$ as the denominator. Looking at the expression, we see that $\frac{3}{n}$ matches this criterion, thus $\frac{3}{n}=\Delta x=\frac{b-a}{n}\implies b-a=3$. Now looking at inside the parentheses, we search for $x_i$, which is equal to $a+(\Delta x)i$. Seeing that $1+\frac{3i}{n}$ fits this form, we can deduce that $a=1$. Thus, we can begin writing our integral. We know that $a$ is 1, and that the difference between $a$ and $b$ is 3, so then we can conclude that $b=4$. So far, we can write this sum as the following: $\displaystyle\int_1^4f(x)dx$. All we have to do now is find $f(x)$. This is an expression that contains $x_i$ within it, so rewriting the interior of the sum as $2(x_i)+1$ shows that $f(x)=2x+1$. We can now fully rewrite our integral: $\displaystyle\lim_{n \to \infty}\sum_{i=1}^n\left[2\left(1+\frac{3i}{n}\right)+1\right]\frac{3}{n}=\displaystyle\int_1^42x+1dx$. Evaluating, we get $x^2+x\biggr\rvert_1^4=20-2=18$.

  2. If we let $h(x)$ be $\displaystyle\int_0^xe^{t^4}$ and $g(x)$ be $3x$, we can rewrite this integral as $h(g(x))$. We know that to take the derivative of a composite function, we need to use the chain rule, that is $\frac{d}{dx}h(g(x))=h'(g(x))g'(x)$. Deriving using the second fundamental theorem of calculus, we get that $F'(x)=3e^{3x^4}$ (in other words, plug $3x$ in for $t$, and then multiply the result by the derivative of $3x$). Evaluating at $x=0$, we get our answer of 3.

  3. We can rewrite this integral as $\displaystyle\int_{2x}^c\cos tdt+\displaystyle\int_c^{5x}\cos t dt=-\displaystyle\int^{2x}_c\cos tdt+\displaystyle\int_c^{5x}\cos t d$, where $c$ is some arbitrary point between $2x$ and $5x$. Now finding the derivative using the chain rule and the second fundamental theorem of calculus, we get our answer: $-2\cos2x+5\cos5x$.

  4. This looks scary, but it’s really not so bad. As before, we must split this integral with some constant $c$ that is between the endpoints. Thus, this integral becomes $\displaystyle\int_{x^4+e^x}^c(t^2-2)dt+\displaystyle\int_c^{(x+1)^2+\sin{x}}(t^2-2)dt=$ $-\displaystyle\int^{x^4+e^x}_c(t^2-2)dt+\displaystyle\int_c^{(x+1)^2+\sin{x}}(t^2-2)dt$. Evaluating this integral using the second fundamental theorem and the chain rule, we plug the function in the bound in for $t$ and multiply the result by the derivative of the function. Doing so, we get the derivative of this integral: $-(((x^4+e^x)^2-2)(4x^3+e^x))+$ $((((x+1)^2+\sin x)^2-2)(2(x+1)+\cos x))$. Evaluating this at $x=0$, we get $1+(-1)(2+1)=1-3=2$.

  5. This can be rewritten as a limit: $\displaystyle\lim_{t \to \infty}\int_0^txe^{-x^2}dx$. Integrating using u-substitution with $u=-x^2$ and $du=-2x$, we get $\displaystyle\lim_{t \to \infty}-\frac{1}{2}\int_0^te^udu$. So, this integral becomes $\displaystyle\lim_{t \to \infty}-\frac{e^u}{2}\biggr\rvert_0^t=\displaystyle\lim_{t \to \infty}-\frac{e^{-x^2}}{2}\biggr\rvert_0^t=\displaystyle\lim_{t \to \infty}-\frac{e^{t^2}}{2}-\displaystyle\lim_{t \to \infty}-\frac{1}{2}=$ $0+\frac{1}{2}=\frac{1}{2}$.

  6. We can deduce that $\displaystyle\int_0^6f(x)dx=4+4=8$. Since we know that $\displaystyle\int_0^6f(x)dx-\displaystyle\int_2^6f(x)dx=\displaystyle\int_0^2f(x)dx$, we can simply do 4+4-5=8 to get our answer.

  7. The answer is $-(a-b)=b-a$.