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Section 5.8 : Substitution Rule for Definite Integrals
Evaluate each of the following integrals, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.
- \( \displaystyle \int_{0}^{1}{{3\left( {4x + {x^4}} \right){{\left( {10{x^2} + {x^5} - 2} \right)}^6},円dx}}\) Solution
- \( \displaystyle \int_{0}^{{\frac{\pi }{4}}}{{\frac{{8\cos \left( {2t} \right)}}{{\sqrt {9 - 5\sin \left( {2t} \right)} }},円dt}}\) Solution
- \( \displaystyle \int_{\pi }^{0}{{\sin \left( z \right){{\cos }^3}\left( z \right),円dz}}\) Solution
- \( \displaystyle \int_{1}^{4}{{\sqrt w ,円{{\bf{e}}^{1 - \sqrt {{w^{,3円}}} }},円dw}}\) Solution
- \( \displaystyle \int_{{ - 4}}^{{ - 1}}{{\sqrt[3]{{5 - 2y}} + \frac{7}{{5 - 2y}},円dy}}\) Solution
- \( \displaystyle \int_{{ - 1}}^{2}{{{x^3} + {{\bf{e}}^{\frac{1}{4}x}},円dx}}\) Solution
- \( \displaystyle \int_{\pi }^{{\frac{{3\pi }}{2}}}{{6\sin \left( {2w} \right) - 7\cos \left( w \right)dw}}\) Solution
- \( \displaystyle \int_{1}^{5}{{\frac{{2{x^3} + x}}{{{x^4} + {x^2} + 1}} - \frac{x}{{{x^2} - 4}},円dx}}\) Solution
- \( \displaystyle \int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} + \frac{3}{{{{\left( {6t - 1} \right)}^2}}},円dt}}\) Solution
- \( \displaystyle \int_{{ - 2}}^{1}{{{{\left( {2 - z} \right)}^3} + \sin \left( {\pi z} \right){{\left[ {3 + 2\cos \left( {\pi z} \right)} \right]}^3},円dz}}\) Solution
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Page Last Modified : 11/16/2022