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Section 7.8 : Improper Integrals
Determine if each of the following integrals converge or diverge. If the integral converges determine its value.
- \( \displaystyle \int_{0}^{\infty }{{\left( {1 + 2x} \right){{\bf{e}}^{ - x}}\,dx}}\) Solution
- \( \displaystyle \int_{{ - \infty }}^{0}{{\left( {1 + 2x} \right){{\bf{e}}^{ - x}}\,dx}}\) Solution
- \( \displaystyle \int_{{ - 5}}^{1}{{\frac{1}{{10 + 2z}}\,dz}}\) Solution
- \( \displaystyle \int_{1}^{2}{{\frac{{4w}}{{\sqrt[3]{{{w^2} - 4}}}}\,dw}}\) Solution
- \( \displaystyle \int_{{ - \infty }}^{1}{{\sqrt {6 - y} \,dy}}\) Solution
- \( \displaystyle \int_{2}^{\infty }{{\frac{9}{{{{\left( {1 - 3z} \right)}^4}}}\,dz}}\) Solution
- \( \displaystyle \int_{0}^{4}{{\frac{x}{{{x^2} - 9}}\,dx}}\) Solution
- \( \displaystyle \int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}}\) Solution
- \( \displaystyle \int_{1}^{4}{{\frac{1}{{{x^2} + x - 6}}\,dx}}\) Solution
- \( \displaystyle \int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}}\) Solution