Integration
Write your working on paper. Reveal each answer only after you've had a go.
1. Integrate the following expressions (give the indefinite integral, and don't forget + C).
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\(I = \int \left(x^{4} - 5x^{3} - 4x^{2} + 3x + 5\right)\,dx\)
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Reverse the power rule: add one to each exponent and divide by the new exponent.\(I = \frac{1}{5}x^{5} - \frac{5}{4}x^{4} - \frac{4}{3}x^{3} + \frac{3}{2}x^{2} + 5x + C\) -
\(I = \int 2\sin x\,dx\)
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The integral of sin is −cos.\(I = -2\cos x + C\) -
\(I = \int -\cos x\,dx\)
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The integral of cos is sin.\(I = -\sin x + C\) -
\(I = \int -2e^{x}\,dx\)
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e raised to x integrates to itself.\(I = -2e^{x} + C\)
2. Integrate using the reverse chain rule.
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\(I = \int 5\sin\left(3x + 8\right)\,dx\)
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Let \(u = 3x + 8\), so \(du/dx = 3\); divide by 3 when you integrate.\(I = -\frac{5}{3}\cos\left(3x + 8\right) + C\) -
\(I = \int -2\cos\left(4x + 3\right)\,dx\)
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Let \(u = 4x + 3\), so \(du/dx = 4\); divide by 4 when you integrate.\(I = -\frac{1}{2}\sin\left(4x + 3\right) + C\) -
\(I = \int 7e^{8x + 4}\,dx\)
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Let \(u = 8x + 4\), so \(du/dx = 8\); divide by 8 when you integrate.\(I = \frac{7}{8}e^{8x + 4} + C\) -
\(I = \int \frac{8}{5x - 1}\,dx\)
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Let \(u = 5x - 1\), so \(du/dx = 5\); divide by 5 when you integrate.\(I = \frac{8}{5}\ln\left|5x - 1\right| + C\)
3. Evaluate the following definite integrals.
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\(I = \int_{1}^{3} \left(6x^{2} + 7x + 8\right)\,dx\)
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Integrate to \(2x^{3} + \frac{7}{2}x^{2} + 8x\), then evaluate at x=3 minus at x=1.\(I = 96\) -
\(I = \int_{0}^{1.5} \sin\left(2x\right)\,dx\)
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Antiderivative \(-\frac{1}{2}\cos\left(2x\right)\); evaluate between the limits.\(I = 0.9950\) -
\(I = \int_{0}^{1.0} \cos\left(x + 1\right)\,dx\)
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Antiderivative \(\frac{1}{1}\sin\left(x + 1\right)\); evaluate between the limits.\(I = 0.06783\) -
\(I = \int_{0}^{1} e^{x}\,dx\)
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Antiderivative \(\frac{1}{1}e^{x}\); evaluate between the limits.\(I = 1.718\)