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Section 4.4 #10: Compute $$\displaystyle\int_1^2 6x^2-3x \rm{d}x.$$
Solution: Calculate $$\begin{array}{ll} \displaystyle\int_1^2 6x^2 -3x \rm{d} x &= \left[ 2x^3 - \dfrac{3}{2} x^2 \right|_1^2 \\ &= \left( 2 \cdot 8 - \dfrac{3}{2} 2^2 \right) - \left( 2 - \dfrac{3}{2} \right) \\ &= 10 - \dfrac{1}{2} \\ &= \dfrac{19}{2}. \end{array}$$

Section 4.4 #17: Compute $$\displaystyle\int_{-1}^1 \sqrt[3]{t}-2 \rm{d}t.$$
Solution: Calculate $$\begin{array}{ll} \displaystyle\int_{-1}^1 \sqrt[3]{t}-2 \rm{d}t &= \displaystyle\int_{-1}^1 t^{\frac{1}{3}} - 2 \rm{d} t \\ &= \left[ \dfrac{t^{\frac{4}{3}}}{\frac{4}{3}} - 2t \right|_{-1}^1 \\ &= \left( \dfrac{3}{4} - 2 \right) - \left( \dfrac{3}{4} + 2 \right) \\ &= -4. \end{array}$$

Section 4.4 #27: Compute $$\displaystyle\int_0^{\pi} 1+\sin(x) \rm{d}x.$$
Solution: Calculate $$\begin{array}{ll} \displaystyle\int_0^{\pi} 1+\sin(x) \rm{d}x &= \Bigg[ x - \cos(x) \Bigg|_0^{\pi} \\ &= \left( \pi - \cos(\pi) \right) - \left( 0 - \cos(0) \right) \\ &= (\pi - (-1)) - (-1) \\ &=\pi + 2. \end{array}$$

Section 4.4 #45: Find the value(s) of $c$ guaranteed to exist by the mean value theorem for integrals applied to $f(x)=x^3$ over the interval $[0,3]$.
Solution: We need to find the value(s) $c$ with $0 \leq c \leq 3$ that obey the formula $$\displaystyle\int_0^3 x^3 \rm{d}x = c^3 (3-0).$$ First compute the left-hand side to get $$\displaystyle\int_0^3 x^3 \rm{d}x = \dfrac{x^4}{4} \Bigg|_0^3 = \dfrac{3^4}{4} = \dfrac{81}{4}.$$ Therefore the value of $c$ we must find must satisfy $$\dfrac{81}{4} = 3c^3,$$ in other words, $$\dfrac{81}{12} = c^3,$$ so taking the cube root, $$\sqrt[3]{\dfrac{81}{12}}=c.$$ The numerator could be simplified, since $\sqrt[3]{81}=3\sqrt[3]{3}$, but this is not necessary.

Section 4.4 #51: Find the average value of $f(x)=9-x^2$ over the interval $[-3,3]$.
Solution: Calculate $$\begin{array}{ll} \dfrac{1}{3-(-3)} \displaystyle\int_{-3}^3 9-x^2 \rm{d}x &= \dfrac{1}{6} \Bigg[ 9x-\dfrac{x^3}{3} \Bigg|_{-3}^3 \\ &=\dfrac{1}{6} \left[ \left( 27 - \dfrac{27}{3} \right) - \left( -27 - \dfrac{-27}{3} \right) \right] \\ &= \dfrac{1}{6} [ (27-9) - (-27+9)] \\ &= \dfrac{18+18}{6} \\ &= \dfrac{36}{6} \\ &= 6. \end{array}$$

Section 4.4 #88: Find $F'(x)$ where $$F(x) = \displaystyle\int_{-x}^x t^3 \rm{d}t.$$
Solution: First split this integral up around some constant, say $0$: $$\begin{array}{ll} F(x)=\displaystyle\int_{-x}^x t^3 \rm{d}t &= \displaystyle\int_0^x t^3 \rm{d}t + \displaystyle\int_{-x}^0 t^3 \rm{d}t \\ &= \displaystyle\int_0^x t^3 \rm{d}t - \displaystyle\int_0^{-x} t^3 \rm{d}t. \end{array}$$ If we let $G(x) = \displaystyle\int_0^x t^3 \rm{d}t$ then we know from the fundamental theorem of calculus that $G'(x)=x^3$ and moreover according to the above calculation, we can write $$F(x) = G(x) - G(-x).$$ Therefore we see (using the chain rule to compute the derivative of $G(-x)$) $$F'(x) = \dfrac{\rm{d}}{\rm{d}x} G(x) - \dfrac{\rm{d}}{\rm{d}x} G(-x) = G'(x) + G'(-x)=x^3 + (-x)^3 = 0.$$

Section 4.5 #11: Find the indefinite integral: $$\displaystyle\int x^2(x^3-1)^4 \rm{d}x.$$
Solution: Let $u=x^3-1$ then $\rm{d}u=3x^2 \rm{d}x$ and so $\dfrac{1}{3} \rm{d}u = x^2 \rm{d}x$. Therefore using substitution, we get $$\begin{array}{ll} \displaystyle\int x^2 (x^3-1)^4 \rm{d}x &= \dfrac{1}{3} \displaystyle\int u^4 \rm{d}u \\ &= \dfrac{1}{15} u^5 +C \\ &= \dfrac{1}{15} (x^3-1)^5 + C. \end{array}$$

Section 4.5 #42: Find the indefinite integral: $$\displaystyle\int \dfrac{\sin(x)}{\cos^3(x)} \rm{d}x.$$
Solution: Let $u=\cos(x)$ then $\rm{d}u = -\sin(x) \rm{d}x$ and so $-\rm{d}u=\sin(x)\rm{d}x$. Therefore we compute $$\begin{array}{ll} \displaystyle\int \dfrac{\sin(x)}{\cos^3(x)} \rm{d}x &= -\displaystyle\int \dfrac{1}{u^3} \rm{d}u \\ &= - \displaystyle\int u^{-3} \rm{d}u \\ &= -\dfrac{u^{-2}}{-2} + C \\ &= \dfrac{1}{2\cos^2(x)} + C. \end{array}$$

Section 4.5 #52: Find the indefinite integral: $$\displaystyle\int \dfrac{2x+1}{\sqrt{x+4}} \rm{d}x.$$
Solution: Let $u=x+4$ so that $\rm{d}u = \rm{d}x$. Also $x=u-4$ and so $2x=2u-8$ and $2x+1=2u-7$. Substituting these values yields $$\begin{array}{ll} \displaystyle\int \dfrac{2x+1}{\sqrt{x+4}} \rm{d}x &= \displaystyle\int \dfrac{2u-7}{\sqrt{u}} \rm{d}u \\ &= 2\displaystyle\int u^{\frac{1}{2}} \rm{d}u - 7 \displaystyle\int u^{-\frac{1}{2}} \rm{d}u \\ &= 2 \left[ \dfrac{u^{\frac{3}{2}}}{\frac{3}{2}} \right] - 7 \left[ \dfrac{u^{\frac{1}{2}}}{\frac{1}{2}} \right] + C \\ &= \dfrac{4}{3} (x+4)^{\frac{3}{2}} - 14 (x+4)^{\frac{1}{2}} + C. \end{array}$$

Section 4.5 #56: Compute $$\displaystyle\int_0^1 x^3 (2x^4+1)^2 \rm{d}x.$$
Solution: Let $u=2x^4+1$ so $\rm{d}u=8x^3 \rm{d}x$ and $\dfrac{1}{8} \rm{d}u = x^3 \rm{d}x$. Also if $x=0$ then $u=2(0^4)+1=1$ and if $x=1$ then $u=2(1^4)+1=3$. Therefore $$\begin{array}{ll} \displaystyle\int_0^1 x^3 (2x^4+1)^2 \rm{d}x &= \dfrac{1}{8} \displaystyle\int_1^3 u^2 \rm{d}u \\ &= \dfrac{1}{8} \left[ \dfrac{u^3}{3} \right|_1^3 \\ &=\dfrac{1}{8} \left[ \dfrac{27}{3} - \dfrac{1}{3} \right] \\ &= \dfrac{1}{8} \left( \dfrac{26}{3} \right) \\ &= \dfrac{13}{12}. \end{array}$$

Section 4.5 #60: Compute $$\displaystyle\int_0^2 \dfrac{x}{\sqrt{1+2x^2}} \rm{d}x.$$
Solution: Let $u=1+2x^2$ so $\rm{d}u=4x \rm{d}x$ and $\dfrac{1}{4} \rm{d}u = \rm{d}x$. Also if $x=0$ then $u=1$ and if $x=2$ then $u=9$. Compute $$\begin{array}{ll} \displaystyle\int_0^2 \dfrac{x}{\sqrt{1+2x^2}} \rm{d}x &= \dfrac{1}{4} \displaystyle\int_1^9 u^{-\frac{1}{2}} \rm{d}u \\ &= \dfrac{1}{4} \left[ \dfrac{u^{\frac{1}{2}}}{\frac{1}{2}} \right|_1^9 \\ &= \dfrac{1}{2} \left[ \sqrt{9} - \sqrt{1} \right] \\ &= \dfrac{2}{2} \\ &= 1. \end{array}$$