5.3 The Fundamental Theorem of Calculus/27: Difference between revisions

Revision as of 21:15, 6 September 2022

<math> \begin{align}

\int_0^2 x(2+x^5)\,dx = \int_0^2 (2x+x^6)\,dx &= \int_0^2 (2x+x^6)\,dx \\[2ex]

&= \left(\frac{2x^{1+1}}{1+1}+\frac{x^{6+1}}{6+1}\right)\bigg|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg|_{0}^{2} \\[2ex]

&= \left((2)^2-\frac{(2)^7}{7}\right)+\left((0)^2+\frac{(0)^7}{7}\right) \\[2ex]

&= \left[4+\frac{2^7}{7}\right]-[0] \\[2ex] &= \frac{156}{7}

\end{align}

@@ Line 2: / Line 2: @@
 \begin{align}
-\int_2^0 x(2+x^5)\,dx = \int_2^0 (2x+x^6)\,dx &= -\int_0^2 (2x+x^6)\,dx \\[2ex]
+\int_0^2 x(2+x^5)\,dx = \int_0^2 (2x+x^6)\,dx &= \int_0^2 (2x+x^6)\,dx \\[2ex]
-&= \left(\frac{2x^2}{1+1}+\frac{x^6+1}{6+1}\right)\bigg|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg|_{0}^{2}
+&= \left(\frac{2x^{1+1}}{1+1}+\frac{x^{6+1}}{6+1}\right)\bigg|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg|_{0}^{2} \\[2ex]
-&= \left((2)^2+\frac{(2)^7}{7}\right)-\left((0)^2+\frac{0^7}{7}\right)
-&= 4+\frac{2^7}{7}
+&= \left((2)^2-\frac{(2)^7}{7}\right)+\left((0)^2+\frac{(0)^7}{7}\right) \\[2ex]
-&= \frac{156}{7}
+&= \left[4+\frac{2^7}{7}\right]-[0] \\[2ex]
+&= \frac{156}{7}
 \end{align}
-</math>