sending the attached document , the error is on page 37 of adobe, the syntax of the set of equations is the following thank you in advance Help
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\item[3c] Si $p=\infty$, entonces $q=1$ y se tiene lo siguiente\begin{align*}\int_{a}^{b}|s_4(x)|\,dx &= -\frac{1}{24}\int_{a}^{\frac{a+b}{2}} (x-a)^3\left(x-\frac{a+2b}{3}\right)\,dx - \frac{1}{24}\int_{\frac{a+b}{2}}^{b} (x-b)^3\left(x-\frac{2a+b}{3}\right)\,dx\\&=-\frac{1}{24}\left[\int_{a}^{\frac{a+b}{2}}(x^3-3ax^2+3a^2x-a^3)\left(x-\frac{a+2b}{3}\right)\,dx\right.\\&\left.+\int_{\frac{a+b}{2}}^{b}(x^3-3bx^2+3b^2x-b^3)\left(x-\frac{2a+b}{3}\right)\,dx\right]\\&=-\frac{1}{24}\left[\int_{a}^{\frac{a+b}{2}}\left[x^4-3ax^3+3a^2x^2-a^3x-x^3\left(\frac{a+2b}{3}\right)\right.\right.\\&\left.+ax^2(a+2b)-a^2x(a+2b)+\frac{a^3}{3}(a+2b)\right]\,dx\\+\left.\int_{\frac{a+b}{2}}^{b}\left[x^4-3bx^3+3b^2x^2-b^3x - \frac{x^3}{3}(2a+b)+bx^2(2a+b)-b^2x(2a+b)+\frac{b^3}{3}(2a+b)\right]\,dx\right]\\&=-\frac{1}{24}\left[\frac{x^5}{5}-\frac{3}{4}ax^4+a^2x^3-\frac{a^3x^2}{2}-\frac{x^4}{12}(a+2b)+\frac{ax^3}{3}(a+2b)-\frac{a^2x^2}{2}\right.\\&\left.\left.+\frac{a^3}{3}(a+2b)x\right]\right|_a^{\frac{a+b}{2}}-\frac{1}{24}\left[\frac{x^5}{5}-\frac{3}{4}bx^4+b^2x^3-\frac{b^3x^2}{2}-\frac{x^4}{12}(2a+b)\right.\\&\left.\left.+\frac{bx^3}{3}(2a+b)-\frac{b^2x^2}{2}+\frac{b^3}{3}(2a+b)x\right]\right|_{\frac{a+b}{2}}^{b}\\&=-\frac{1}{24}\left[\frac{1}{5}\left(\frac{a+b}{2}\right)^5-\frac{3a}{4}\left(\frac{a+b}{2}\right)^4+a^2\left(\frac{a+b}{2}\right)^3-\frac{a^3}{2}\left(\frac{a+b}{2}\right)^2\right.\\&-\frac{1}{12}\left(\frac{a+b}{2}\right)^4(a+2b)+\frac{a}{3}\left(\frac{a+b}{2}\right)^3(a+2b)-\frac{a^2}{2}\left(\frac{a+b}{2}\right)^2(a+2b)\\&+\frac{a^3}{3}(a+2b)\left(\frac{a+b}{2}\right)-\frac{a^5}{5}+\frac{3}{4}a^5-a^5+\frac{a^5}{2}+\frac{a^4}{12}(a+2b)-\frac{a^4}{3}(a+2b)\\&+\frac{a^4}{2}(a+2b)-\frac{a^4}{3}(a+2b)\\&+\frac{b^5}{5}-\frac{3}{4}b^5+b^5-\frac{b^5}{2}^-\frac{b^4}{12}(2a+b)+\frac{b^4}{3}(2a+b)-\frac{b^4}{2}(2a+b)+\frac{b^4}{3}(2a+b)\\&-\frac{1}{5}\left(\frac{a+b}{2}\right)^5+\frac{3b}{4}\left(\frac{a+b}{2}\right)^4-b^2\left(\frac{a+b}{2}\right)^3+\frac{b^3}{2}\left(\frac{a+b}{2}\right)^2\\&\left.+\frac{1}{12}\left(\frac{a+b}{2}\right)^4(2a+b)-\frac{b}{3}\left(\frac{a+b}{2}\right)^3(2a+b)+\frac{b^2}{2}\left(\frac{a+b}{2}\right)^2(2a+b)-\frac{b^3}{3}(2a+b)\left(\frac{a+b}{2}\right)\right]\\&=-\frac{1}{24}\left[\frac{b^5-a^5}{20}+\frac{1}{12}(b^5-a^5+2ab^4-2ba^4)+2\left(\frac{a+b}{2}\right)^4(b-a)\right.\\&\left.+\frac{4}{3}(a^2-b^2)\left(\frac{a+b}{2}\right)^3+\left(\frac{a+b}{2}\right)^2[b^3-a^3+b^2a-a^2b]+\left(\frac{a+b}{2}\right)\left(\frac{a^3}{3}(a+2b)-\frac{b^3}{3}(2a+b)\right)\right]\\\end{align*}
