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Section 4.4

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1. (a) \displaystyle{\int (x^3 + 3x - 6)dx = \frac{1}{4}x^4 + \frac{3}{2}x^2 - 6x + c}

(c) \displaystyle{\int \frac{1}{x^4} \ dx = -\frac{1}{3x^3} + c}

(e) \displaystyle{\int \frac{12}{\sqrt{t}} \ dt = 24\sqrt{t} + c}

(g) \displaystyle{\int 7\sqrt{x + 5} \ dx = \frac{14}{3}(x + 5)^{\frac{3}{2}} + c}

2. (a) \displaystyle{\int \sin(3x)dx = -\frac{1}{3}\cos(3x) + c}

(c) \displaystyle{\int \sqrt{3t - 1} \ dt = \frac{2}{9}(3t - 1)^{\frac{3}{2}} + c}

(e) \displaystyle{\int 7\sec^2(2x)dx = \frac{7}{2}\tan(2x) + c}

(g) \displaystyle{\int 2\csc^2(7x)dx = -\frac{2}{7}\cot(7x) + c}

3. \displaystyle{\int 6x\sqrt{1 + 3x^2} \ dx = \frac{2}{3}(1 + 3x^2)^{\frac{3}{2}} + c}

(c) \displaystyle{\int x^2(3 + x^3)^{10} dx = \frac{1}{33}(3 + x^3)^{11} + c}

(e) \displaystyle{\int 4t\sin(t^2)dt = -2\cos(t^2) + c}

(g) \displaystyle{\int \sin^3(t)\cos(t)dt = \frac{1}{4}\sin^4(t) + c}

4. (a) \displaystyle{\int \frac{\sin(\sqrt{x})}{\sqrt{x}} \ dx = -2\cos(\sqrt{x}) + c}

(c) \displaystyle{\int \sec^3(4x)\tan(4x)dx = \frac{1}{12}\sec^3(4x) + c}

(e) \displaystyle{\int \frac{\sin(x)}{\cos^2(x)} \ dx = \sec(x) + c}

(f) \displaystyle{\int t\sqrt{t - 2} \ dt = \frac{2}{5}(t - 2)^{\frac{5}{2}} + \frac{4}{3}(t - 2)^{\frac{3}{2}} + c}

5. \displaystyle{\int_0^1 (4x^2 - 3x - 5)dx = -\frac{31}{6}}

(c) \displaystyle{\int_0^{\frac{\pi}{4}} 3\sin(2x))dx = \frac{3}{2}}

(e) \displaystyle{\int_{-\frac{\pi}{12}}^0 6\sec(3t)\tan(3t)dt = 2 - 2\sqrt{2}}

(g) \displaystyle{\int_0^2 x\sqrt{x^2 + 1} \ dx = \frac{5\sqrt{5} - 1}{3}}

6. (a) \displaystyle{\int_{-1}^1 \frac{5x^2}{(x^3 + 2)^2} \ dx = \frac{10}{9}}

(c) \displaystyle{\int_0^{\sqrt{\pi}} 3x\sin(x^2)dx = 3}

(e) \displaystyle{\int_0^{\frac{\pi}{2}} \sin^3(2t)\cos(2t)dt = 0}

(g) \displaystyle{\int_{-1}^1 x(1 + x^2)^{25} dx = 0}

7. (a) \displaystyle{\int_0^3 \frac{x}{\sqrt{x+1}} \ dx = \frac{8}{3}}

(c) \displaystyle{\int_{-\frac{\pi}{12}}^{\frac{\pi}{24}} \tan^4(4x)\sec^2(4x)dx = \frac{61}{45\sqrt{3}}}

(e) \displaystyle{\int_0^1 4x(1 + x)^{25}dx = \frac{3,355,443,202}{351}}

(g) \displaystyle{\int_1^5 5u\sqrt{2u - 1} \ du = \frac{428}{3}}

8. Area \displaystyle{= \int_0^{\frac{\pi}{6}} 4\sin(6t)dt = \frac{4}{3}}

This entry was posted on 22 July 22 2007 at 5:25 pm and is filed under Section 4.4. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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