Calculus exercise 7.2

EXERCISE 7.2

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Integration by parts technique is used to solve problems in a simple way with complete steps.

1-38 Evaluate the integral.

  1.     \[\int x e^{-2 x} d x\]

Integrating by parts

    \[\int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x\]

Here u=x, v=e^{-2 x}

    \[\begin{aligned} \therefore \quad \int x e^{-2 x} d x & =x \int e^{-2 x} d x-\int\left[\frac{d(x)}{d x} \int e^{-2 x} d x\right] d x \\ \\ & =x \frac{e^{-2 x}}{-2}-\int(1) \frac{e^{-2 x}}{-2} d x \\ \\ & =-\frac{x e^{-2x}}{2}+\frac{1}{2} \int e^{-2 x} d x \\ \\ & =-\frac{x e^{-2 x}}{2}+\frac{1}{2}\left[\frac{e^{-2 x}}{-2}\right]+c \\ \\ & =-\frac{x e^{-2 x}}{2}-\frac{1}{4} e^{-2 x}+c \end{aligned}\]

2.

    \[\int x e^{3 x} d x\]

Here

    \[u=x, \quad v=e^{3 x}\]

Formula integration by parts:

    \[\int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x\]

integrating by parts

    \[\begin{aligned} & \int x e^{3 x} d x=x \int e^{3x} d x-\int\left[\frac{d}{d x}(x) \int e^{3 x} d x\right] d x \\ \\ &=x \frac{e^{3x}}{3}-\int(1) \frac{e^{3x}}{3} d x \\ \\ &=\frac{x e^{3x}}{3}-\frac{1}{3} \int e^{3 x} d x \\ \\ &=\frac{x e^{3x}}{3}-\frac{1}{3}\left[\frac{e^{3x}}{3}\right]+c \\ \\ &=\frac{x e^{3x}}{3}-\frac{1}{9} e^{3x}+c \end{aligned}\]

3.

    \[\begin{aligned} & \int x^{2} e^{x} d x \\ \\ & u=x^{2}, \quad v=e^{x} \\ \\ & \text {Formula integration by parts:} \quad \int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x \end{aligned}\]

Integrating by parts

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

Again integrating by parts

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

4.

    \[\int x^{2} e^{-2 x} d x\]

    \[u=x^{2}, \quad v=e^{-2 x}\]

Formula integration by parts:

    \[\int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x\]

Integrating by parts

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

Again integrating by parts

    \[\begin{aligned} & =-\frac{x^{2} e^{-2 x}}{2}+x \int e^{-2 x} d x-\int \left[\frac{d}{d x}\left(x\right) \int e^{-2 x} d x\right]dx \\ \\ &=-\frac{x^{2} e^{-2 x}}{2}+x \frac{e^{-2x}}{-2}- \int (1)\frac{e^{-2x}}{-2} dx \\ \\ &=-\frac{x^{2} e^{-2 x}}{2}- \frac{xe^{-2x}}{2} +\frac{1}{2} \int e^{-2x} dx \\ \\ &=-\frac{x^{2} e^{-2 x}}{2}- \frac{xe^{-2x}}{2} +\frac{1}{2} \left(\frac{e^{-2x}}{-2}\right)+c \\ \\ &=-\frac{x^{2} e^{-2 x}}{2}- \frac{xe^{-2x}}{2}-\frac{1}{4} e^{-2x}+c \end{aligned}\]

5.

    \[\begin{aligned} & \int x \sin 3x d x \\ \\ & u=x, \quad v=\sin 3x \\ \\ & \text {Formula integration by parts:} \quad \int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x \end{aligned}\]

Integrating by parts

    \[\begin{aligned} \int x \sin 3x d x= x \int \sin 3x dx-\int \left [ \frac {d}{dx}(x) \int\sin 3x dx \right]dx \\ \\ =x \left (\frac{-\cos 3x}{3}\right)-\int (1) \left(\frac{-\cos 3x}{3}\right)dx \\ \\ =-\frac{1}{3} x \cos 3x +\frac{1}{3} \int \cos 3x dx \\ \\ =-\frac{1}{3} x \cos 3x +\frac{1}{3} \left(\frac{\sin 3x}{3}\right)+c \\ \\ =-\frac{1}{3} x \cos 3x +\frac{1}{9} \sin 3x+c \end{aligned}\]