Generating...                               quiz06c_n25

  1. Evaluate $\displaystyle\int\!\!\int_D -x\,y dxdy$ where $D$ is the region bounded by $\displaystyle x=y^2$ and x = −3y − 2.

    $\displaystyle -\int_{-2}^{-1}{y\,\left({{9\,y^2+12\,y+4}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle {{5}\over{8}}$ $\displaystyle \int_{-1}^{0}{y\,\left({{9\,y^2+12\,y+4}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle -{{1}\over{24}}$ $\displaystyle -\int_{-1}^{0}{y\,\left({{9\,y^2+12\,y+4}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle {{1}\over{24}}$ $\displaystyle \int_{-2}^{-1}{y\,\left({{9\,y^2+12\,y+4}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle -{{5}\over{8}}$

  2. Evaluate $\displaystyle \int_{1}^{2}{y\,\int_{1}^{2}{x\,\sqrt{y^2+x^2}\;dx}\;dy}$.

    $\displaystyle \int_{1}^{2}{y\,\sqrt{y^2+4}\;dy}$ = $\displaystyle {{2^{{{9}\over{2}}}}\over{3}}-{{5^{{{3}\over{2}}}}\over{3}}$ $\displaystyle \int_{1}^{2}{y\,\left({{\left(y^2+4\right)^{{{3}\over{2}}}}\over{3 }}-{{\left(y^2+1\right)^{{{3}\over{2}}}}\over{3}}\right)\;dy}$ = $\displaystyle -{{5^{{{5}\over{2}}}-2^{{{15}\over{2}}}}\over{15}}-{{5^{{{5}\over{2 }}}-2^{{{5}\over{2}}}}\over{15}}$ $\displaystyle \int_{1}^{2}{y\,\left(y^2+4\right)^{{{3}\over{2}}}\;dy}$ = $\displaystyle {{2^{{{15}\over{2}}}}\over{5}}-5^{{{3}\over{2}}}$ $\displaystyle \int_{1}^{2}{y\,\left(2\,\sqrt{y^2+4}-2\,\sqrt{y^2+1}\right)\;dy}$ = $\displaystyle -{{2\,5^{{{3}\over{2}}}-2^{{{11}\over{2}}}}\over{3}}-{{2\,5^{{{3 }\over{2}}}-2^{{{5}\over{2}}}}\over{3}}$

  3. Evaluate $\displaystyle \int_{0}^{1}{\int_{1}^{2}{e^{x-1}\,y^2+x\;dx}\;dy}$.

    $\displaystyle \int_{0}^{1}{e^2\,y^2-e\,y^2+1\;dy}$ = $\displaystyle {{2\,e+7}\over{6}}$ $\displaystyle \int_{0}^{1}{e^2\,y^2-e\,y^2+1\;dy}$ = $\displaystyle {{2\,e+7}\over{6}}$ $\displaystyle \int_{0}^{1}{-e^2\,y^2+e\,y^2+1\;dy}$ = $\displaystyle {{2\,e+7}\over{6}}$ $\displaystyle {{\int_{0}^{1}{\left(2\,e-2\right)\,y^2+3\;dy}}\over{2}}$ = $\displaystyle {{2\,e+7}\over{6}}$

  4. Evaluate $\displaystyle \int_{0}^{1}{\int_{0}^{1}{e^{y+x}\;dy}\;dx}$.

    $\displaystyle \int_{0}^{1}{x\,e^{x}\;dx}$ = 1 $\displaystyle \int_{0}^{1}{e^{x+1}-e^{x}\;dx}$ = $\displaystyle e^2-2\,e+1$ $\displaystyle \left(e-1\right)\,\int_{0}^{1}{x\;dx}$ = $\displaystyle {{e-1}\over{2}}$ $\displaystyle {{\int_{0}^{1}{e^{x}\;dx}}\over{2}}$ = $\displaystyle {{e-1}\over{2}}$

  5. Evaluate $\displaystyle \int_{0}^{1}{\int_{0}^{1}{\left(y+x\right)^2\;dy}\;dx}$.

    $\displaystyle {{\int_{0}^{1}{3\,x^2+3\,x+1\;dx}}\over{3}}$ = $\displaystyle {{7}\over{6}}$ $\displaystyle \int_{0}^{1}{\left(x+1\right)^2-x^2\;dx}$ = $\displaystyle {{7}\over{6}}$ $\displaystyle \int_{0}^{1}{x^2\;dx}$ = $\displaystyle {{7}\over{6}}$ $\displaystyle {{\int_{0}^{1}{2\,x+1\;dx}}\over{2}}$ = $\displaystyle {{7}\over{6}}$

  6. Evaluate $\displaystyle \int_{1}^{2}{\int_{1}^{4}{{{y+x}\over{\sqrt{x}}}\;dx}\;dy}$.

    $\displaystyle \int_{1}^{2}{{{1}\over{2}}-{{y+4}\over{16}}\;dy}$ = $\displaystyle {{5}\over{32}}$ $\displaystyle {{\int_{1}^{2}{\ln 4\,y+3\;dy}}\over{2}}$ = $\displaystyle {{23}\over{3}}$ $\displaystyle \int_{1}^{2}{{{y+4}\over{2}}-y-1\;dy}$ = $\displaystyle {{5}\over{32}}$ $\displaystyle {{\int_{1}^{2}{6\,y+14\;dy}}\over{3}}$ = $\displaystyle {{23}\over{3}}$

  7. Evaluate $\displaystyle \int_{0}^{{{\pi}\over{4}}}{\int_{0}^{{{\pi}\over{4}}}{\sin \left(y+ x\right)\;dx}\;dy}$.

    $\displaystyle \int_{0}^{{{\pi}\over{4}}}{\cos y\;dy}$ = $\displaystyle {{1}\over{\sqrt{2}}}$ $\displaystyle \left(1-{{1}\over{\sqrt{2}}}\right)\,\int_{0}^{{{\pi}\over{4}}}{ \sin y\;dy}$ = $\displaystyle \left(1-{{1}\over{\sqrt{2}}}\right)^2$ $\displaystyle \int_{0}^{{{\pi}\over{4}}}{\cos y-\cos \left({{4\,y+\pi}\over{4}} \right)\;dy}$ = $\displaystyle \sqrt{2}-1$ $\displaystyle \int_{0}^{{{\pi}\over{4}}}{\sin y\;dy}$ = $\displaystyle 1-{{1}\over{\sqrt{2}}}$

  8. Evaluate $\displaystyle \int_{0}^{\pi}{\int_{0}^{{{\pi}\over{2}}}{3\,\sin y-3\,\cos x\;dy} \;dx}$.

    $\displaystyle -{{\int_{0}^{\pi}{3\,\pi\,\sin x+6\;dx}}\over{2}}$ = $\displaystyle 3\,\pi$ $\displaystyle {{\int_{0}^{\pi}{3\,\pi\,\cos x+6\;dx}}\over{2}}$ = $\displaystyle 3\,\pi$ $\displaystyle {{\int_{0}^{\pi}{3\,\pi\,\sin x-6\;dx}}\over{2}}$ = $\displaystyle 3\,\pi$ $\displaystyle -{{\int_{0}^{\pi}{3\,\pi\,\cos x-6\;dx}}\over{2}}$ = $\displaystyle 3\,\pi$

  9. Evaluate $\displaystyle\int\!\!\int_D y+x dxdy$ where $D$ is the region bounded by $\displaystyle y=2\,x^2$ and $\displaystyle y=x^2+4$.

    $\displaystyle \int_{-3}^{3}{{{x^4+2\,x^3+8\,x^2+8\,x+16}\over{2}}-2\,x^4-2\,x^3 \;dx}$ = $\displaystyle -{{129}\over{5}}$ $\displaystyle \int_{-3}^{3}{-3\,x^4-x^3+8\,x^2+4\,x+16\;dx}$ = $\displaystyle -{{258}\over{5}}$ $\displaystyle \int_{-2}^{2}{-3\,x^4-x^3+8\,x^2+4\,x+16\;dx}$ = $\displaystyle {{1024}\over{15}}$ $\displaystyle \int_{-2}^{2}{{{x^4+2\,x^3+8\,x^2+8\,x+16}\over{2}}-2\,x^4-2\,x^3 \;dx}$ = $\displaystyle {{512}\over{15}}$

  10. Evaluate $\displaystyle \int_{0}^{1}{\int_{0}^{1}{\left(y+4\,x\right)^3\;dx}\;dy}$.

    $\displaystyle \int_{0}^{1}{\left(y+4\right)^4\;dy}$ = $\displaystyle {{2101}\over{5}}$ $\displaystyle \int_{0}^{1}{\left(y+4\right)^3\;dy}$ = $\displaystyle {{369}\over{4}}$ $\displaystyle \int_{0}^{1}{y^3+6\,y^2+16\,y+16\;dy}$ = $\displaystyle {{105}\over{4}}$ $\displaystyle \int_{0}^{1}{{{y^3+12\,y^2+48\,y+64}\over{3}}-{{y^3}\over{3}}\;dy}$ = $\displaystyle {{92}\over{3}}$


Department of Mathematics
Last modified: 2025-12-21