Calculus Solution Ex #5.2

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In this section, we shall develop the concept of antiderivatives. It contains solutions to integration problems.

(1) — In each part, confirm that the formula is correct….

(a)

    \[\begin{aligned} & \frac{d}{d x}\left[\sqrt{1+x^{2}}\right] \\ \\ & =\frac{d}{d x}\left[1+x^{2}\right]^{1 / 2} \\ \\ & =\frac{1}{2}\left[1+x^{2}\right]^{-1 / 2}(2 x) \\ \\ & =\frac{x}{\left(1+x^{2}\right)^{1 / 2}}=\frac{x}{\sqrt{1+x^{2}}} \end{aligned}\]

corresponding integration formula:

    \[\int \frac{x}{\sqrt{1+x^{2}}} d x=\sqrt{1+x^{2}}+c_{1}\]

(b)

    \[\begin{aligned} \frac{d}{d x}\left[x e^{x}\right] & =x e^{x}+e^{x} \\ \\ & =(x+1) e^{x} \end{aligned}\]

Corresponding integration formula:

    \[\int(x+1) e^{x} d x=x e^{x}+c_{1}\]

 

(2)—In each part, confirm that the stated formula is correct by differentiating.

(a)

    \[\begin{aligned} & \int x \sin x d x=\sin x-x \cos x+c \\ \\ & \text { We shall take derivative of } \quad \sin x-x \cos x+c \\ \\ & \frac{d}{d x}[\sin x-x \cos x+c]=\cos x-\cos x-x(-\sin x) \\ \\ &=x \sin x \end{aligned}\]

(b)

    \[\begin{aligned} \int \frac{d x}{\left(1-x^{2}\right)^{3 / 2}} & =\frac{x}{\sqrt{1-x^{2}}}+c \\ \\ & \text { We shall now differentiate} \quad \frac{x}{\sqrt{1-x^{2}}}+c \\ \\ & \frac{d}{d x}\left[\frac{x}{\sqrt{1-x^{2}}}+c\right] =\frac{d}{d x}\left[x\left(1-x^{2}\right)^{-1 / 2}+c\right] \\ \\ & =(1)\left(1-x^{2}\right)^{-3 / 2}+x\left(1-x^{2}\right)^{-3 / 2}(-2 x)\left(\frac{1}{2}\right)+0 \\ \\ & =\frac{1}{\left(1-x^{2}\right)^{1 / 2}}+\frac{x^{2}}{\left(1-x^{2}\right)^{3 / 2}} \\ \\ & =\frac{1-x^{2}+x^{2}}{\left(1-x^{2}\right)^{3 / 2}}=\frac{1}{\left(1-x^{2}\right)^{3 / 2}} \end{aligned}\]

5-8. Find the derivative and state a corresponding integration formula.

(5).

    \[\frac{d}{d x}\left[\sqrt{x^{3}+5}\right]\]

 

    \[\begin{aligned} \frac{d}{d x}\left[x^{3}+5\right]^{1 / 2} & =\frac{1}{2}\left[x^{3}+5\right]^{-1 / 2}\left(3 x^{2}\right) \\ \\ & =\frac{3 x^{2}}{2\left(x^{3}+5\right)^{1 / 2}} \end{aligned}\]

Corresponding integration formula:

    \[\int \frac{3 x^{2}}{2\left(x^{3}+5\right)^{1 / 2}} d x=\left(x^{3}+5\right)^{1 / 2}+c\]

(6).

    \[\begin{aligned} \frac{d}{d x}\left[\frac{x}{x^{2}+3}\right] & =\frac{\left(x^{2}+3\right)(1)-x(2 x)}{\left(x^{2}+3\right)^{2}} \\ \\ & =\frac{x^{2}+3-2 x^{2}}{\left(x^{2}+3\right)^{2}} \\ & =\frac{3-x^{2}}{\left(x^{2}+3\right)^{2}} \end{aligned}\]

Corresponding integration formula

    \[\int \frac{3-x^{2}}{\left(x^{2}+3\right)^{2}} d x=\frac{x}{x^{2}+3}+c\]

(7).

    \[\begin{aligned} \frac{d}{d x}[\sin (2 \sqrt{x})] & =\cos (2 \sqrt{x}) \frac{d}{d x}(2 \sqrt{x}) . \\ \\ & =2 \cos (2 \sqrt{x}) \cdot \frac{1}{2} x^{-1 / 2} \\ \\ & =\frac{\cos (2 \sqrt{x})}{\sqrt{x}} \end{aligned}\]

Corresponding integration formula

    \[\begin{aligned} \int \frac{\cos (2 \sqrt{x})}{\sqrt{x}} d x & =\sin (2 \sqrt{x})+c \end{aligned}\]

(8).

    \[\begin{aligned} \frac{d}{d x}[\sin x-x \cos x] & =\cos x-\cos x-x(-\sin x) \\ \\ & =x \sin x . \end{aligned}\]

Corresponding integration formula.

    \[\int x \sin x d x=\sin x-x \cos x+c\]

Power rule:

    \[\int x^{r} d x=\frac{x^{r+1}}{r+1}+c \quad (r \neq-1)\]

9(a)

    \[\int x^{8} d x=\frac{x^{9}}{9}+c_{1}\]

(b)

    \[\begin{aligned} \int x^{5 / 7} d x & =\frac{x^{\frac{5}{7}+1}}{\frac{5}{7}+1}+c_{1} \\ \\ & =\frac{x^{12 / 7}}{12 / 7}+c_{1}=\frac{7 x^{12 / 7}}{12}+c_{1} \end{aligned}\]

(c)

    \[\begin{aligned} \int x^{3} \sqrt{x} d x & =\int x^{3} x^{\frac{1}{2}} d x=\int x^{7 / 2}dx \\ & =\frac{x^{\frac{7}{2}+1}}{\frac{7}{2}+1}+c_{1} \\ \\ & =\frac{x^{9 / 2}}{\frac{9}{2}}+c_{1}=\frac{2 x^{9 / 2}}{9}+c_{1} \end{aligned}\]

10(a)

    \[\int \sqrt[3]{x^{2}} d x=\int x^{2 / 3} d x\]

    \[= \frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}+c_{1}=\frac{3}{5} x^{5 / 3}+c_{1}\]

 

 

10(b)

    \[\begin{aligned} \int \frac{1}{x^{6}} d x=\int x^{-6} d x \\ \\ =\frac{x^{-6+1}}{-6+1}+c_{1}=-\frac{x^{-5}}{-5}+c_{1} \\ \\ =\frac{x^{-5}}{5}+c_{1} \end{aligned}\]

10(c)

    \[\int x^{-7 / 8} d x=\frac{x^{\frac{-7}{8}+1}}{\frac{-7}{8}+1}+c_{1}\]

    \[=\frac{x^{1 / 8}}{1 / 8}+c_{1}=8 x^{1 / 8}+c_{1}\]

(11)

    \[\begin{aligned} \int\left[5 x+\frac{2}{3 x^{5}}\right] d x & =\int 5 x d x+\int \frac{2}{3 x^{5}} d x \\ \\ & =5 \int x d x+\frac{2}{3} \int x^{-5} d x \\ \\ & =5 \frac{x^{2}}{2}+\frac{2}{3} \frac{x^{-5+1}}{(-5+1)}+c_{1}=\frac{5}{2} x^{2}+\frac{2}{3} \frac{x^{-4}}{(-4)}+c_{1} \\ \\ & =\frac{5}{2} x^{2}-\frac{1}{6} x^{-4}+c_{1} \end{aligned}\]

(12).

    \[\begin{aligned} & \int\left[x^{-1 / 2}-3 x^{7 / 5}+\frac{1}{9}\right] d x \\ \\ & =\int x^{-\frac{1}{2}} d x-3 \int x^{7 / 5} d x+\frac{1}{9} \int d x \\ \\ & =\frac{x^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}-3 \frac{x^{\frac{7}{5}+1}}{\frac{7}{5}+1}+\frac{1}{9} x+c_{1} \\ \\ & =\frac{x^{1 / 2}}{1 / 2}-3 \frac{x^{12 / 5}}{12 / 5}+\frac{1}{9} x+c_{1} \\ \\ & =2 x^{\frac{1}{2}}-\frac{5}{4} x^{12 / 5}+\frac{1}{9} x+c_{1} \end{aligned}\]

 

(13).

    \[\begin{aligned} & \int\left[x^{-3}-3 x^{1 / 4}+8 x^{2}\right] d x \\ \\ & =\int x^{-3} d x-3 \int x^{1 / 4} d x+8 \int x^{2} d x \\ \\ & =\frac{x^{-3+1}}{-3+1}-3 \frac{x^{\frac{1}{4}+1}}{\frac{1}{4}+1}+8 \frac{x^{2+1}}{2+1}+c \\ \\ & =-\frac{1}{2} x^{-2}-3 \frac{x^{5 / 4}}{5 / 4}+8 \frac{x^{3}}{3}+c \\ \\ & =-\frac{1}{2} x^{-2}-\frac{12}{5} x^{5 / 4}+\frac{8}{3} x^{3}+c \end{aligned}\]

(14).

    \[\begin{aligned} & \int\left[\frac{10}{y^{3 / 4}}-\sqrt[3]{y}+\frac{4}{\sqrt{y}}\right] d y \\ \\ &=10 \int y^{-3 / 4} d y-\int y^{1 / 3} d y+4 \int \frac{1}{y^{1 / 2}} d y \\ \\ & =10 \frac {y^{\frac{-3}{4}+1}}{\frac{-3}{4}+1}-\frac{y^{\frac{1}{3}+1}}{\frac{1}{3}+1}+4 \frac{y^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+c \\ \\ &=10 \frac {y^{\frac{1}{4}}}{\frac{1}{4}}-\frac{y^{\frac{4}{3}}}{\frac{4}{3}}+4 \frac{y^{\frac{1}{2}}}{\frac{1}{2}}+c \\ \\ &=40 y^{\frac{1}{4}}-\frac{3}{4} {y^{\frac{4}{3}}+8 y^{\frac{1}{2}}+c \end{aligned}\]