Generating...                               quiz06c_n1

  1. Evaluate $\displaystyle \int_{0}^{1}{\int_{0}^{1}{\left(3\,y-x^2\right)^2\;dy}\;dx}$.

    $\displaystyle \int_{0}^{1}{x^4\;dx}$ = $\displaystyle {{11}\over{5}}$ $\displaystyle -{{\int_{0}^{1}{2\,x^2-3\;dx}}\over{2}}$ = $\displaystyle {{11}\over{5}}$ $\displaystyle \int_{0}^{1}{x^4-3\,x^2+3\;dx}$ = $\displaystyle {{11}\over{5}}$ $\displaystyle \int_{0}^{1}{\left(3-x^2\right)^2-x^4\;dx}$ = $\displaystyle {{11}\over{5}}$

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

    $\displaystyle \int_{0}^{1}{e^2\,y^2-e\,y^2+1\;dy}$ = $\displaystyle {{2\,e+13}\over{6}}$ $\displaystyle \int_{0}^{1}{e^3\,y^2-e^2\,y^2+1\;dy}$ = $\displaystyle {{2\,e+13}\over{6}}$ $\displaystyle {{\int_{0}^{1}{\left(2\,e-2\right)\,y^2+5\;dy}}\over{2}}$ = $\displaystyle {{2\,e+13}\over{6}}$ $\displaystyle \int_{0}^{1}{-2\,e^3\,y^2+2\,e^2\,y^2+1\;dy}$ = $\displaystyle {{2\,e+13}\over{6}}$

  3. Evaluate $\displaystyle \int_{0}^{1}{\int_{0}^{1}{\left(y+2\,x\right)^3\;dx}\;dy}$.

    $\displaystyle \int_{0}^{1}{\left(y+2\right)^3\;dy}$ = $\displaystyle {{65}\over{4}}$ $\displaystyle \int_{0}^{1}{{{y^3+6\,y^2+12\,y+8}\over{3}}-{{y^3}\over{3}}\;dy}$ = $\displaystyle {{16}\over{3}}$ $\displaystyle \int_{0}^{1}{y^3+3\,y^2+4\,y+2\;dy}$ = $\displaystyle {{21}\over{4}}$ $\displaystyle \int_{0}^{1}{\left(y+2\right)^4\;dy}$ = $\displaystyle {{211}\over{5}}$

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

    $\displaystyle {{\int_{0}^{\pi}{3\,\pi\,\sin x+4\;dx}}\over{2}}$ = $\displaystyle -3\,\pi$ $\displaystyle \int_{0}^{\pi}{\pi\,\cos x-3\;dx}$ = $\displaystyle -3\,\pi$ $\displaystyle -{{\int_{0}^{\pi}{3\,\pi\,\sin x-4\;dx}}\over{2}}$ = $\displaystyle -3\,\pi$ $\displaystyle \int_{0}^{\pi}{-\pi\,\cos x-3\;dx}$ = $\displaystyle -3\,\pi$

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

    $\displaystyle 2\,\int_{0}^{1}{x\;dx}$ = $\displaystyle {{17}\over{3}}$ $\displaystyle \int_{0}^{1}{{{\left(x+4\right)^2}\over{4}}-\left(x+1\right)^2\;dx}$ = $\displaystyle {{17}\over{3}}$ $\displaystyle \int_{0}^{1}{-{{2\,x+1}\over{2}}+4\,x+8\;dx}$ = $\displaystyle {{17}\over{3}}$ $\displaystyle {{\int_{0}^{1}{6\,x+14\;dx}}\over{3}}$ = $\displaystyle {{17}\over{3}}$

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

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

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

    $\displaystyle \int_{-1}^{1}{3\,x^4-x^3-2\,x^2+x-1\;dx}$ = $\displaystyle -{{32}\over{15}}$ $\displaystyle \int_{-3}^{3}{3\,x^4-x^3-2\,x^2+x-1\;dx}$ = $\displaystyle {{1248}\over{5}}$ $\displaystyle \int_{-1}^{1}{-{{x^4-2\,x^3+2\,x^2-2\,x+1}\over{2}}+2\,x^4-2\,x^3 \;dx}$ = $\displaystyle -{{16}\over{15}}$ $\displaystyle \int_{-3}^{3}{-{{x^4-2\,x^3+2\,x^2-2\,x+1}\over{2}}+2\,x^4-2\,x^3 \;dx}$ = $\displaystyle {{624}\over{5}}$

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

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

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

    $\displaystyle {{\int_{1}^{2}{\ln 4\,y-3\;dy}}\over{2}}$ = $\displaystyle -{{5}\over{3}}$ $\displaystyle \int_{1}^{2}{-y+{{y-4}\over{2}}+1\;dy}$ = $\displaystyle -{{11}\over{32}}$ $\displaystyle {{\int_{1}^{2}{6\,y-14\;dy}}\over{3}}$ = $\displaystyle -{{5}\over{3}}$ $\displaystyle \int_{1}^{2}{-{{y-4}\over{16}}-{{1}\over{2}}\;dy}$ = $\displaystyle -{{11}\over{32}}$

  10. Find the volume of the tetrahedron bounded by z = 2 − y, y = x, $x=0$ and $z=0$.

    $\displaystyle {{\int_{0}^{2}{x^2-2\,x\;dx}}\over{2}}$ = $\displaystyle -{{2}\over{3}}$ $\displaystyle {{\int_{0}^{1}{x^2-2\,x\;dx}}\over{2}}$ = $\displaystyle -{{1}\over{3}}$ $\displaystyle {{\int_{0}^{1}{x^2-4\,x+4\;dx}}\over{2}}$ = $\displaystyle {{7}\over{6}}$ $\displaystyle {{\int_{0}^{2}{x^2-4\,x+4\;dx}}\over{2}}$ = $\displaystyle {{4}\over{3}}$


Department of Mathematics
Last modified: 2026-05-20