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

    $\displaystyle -{{\int_{0}^{1}{6\,x-14\;dx}}\over{3}}$ = $\displaystyle {{11}\over{3}}$ $\displaystyle \int_{0}^{1}{{{\left(4-x\right)^2}\over{4}}-\left(1-x\right)^2\;dx}$ = $\displaystyle {{11}\over{3}}$ $\displaystyle \int_{0}^{1}{{{2\,x-1}\over{2}}-4\,x+8\;dx}$ = $\displaystyle {{11}\over{3}}$ $\displaystyle -2\,\int_{0}^{1}{x\;dx}$ = $\displaystyle {{11}\over{3}}$

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

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

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

    $\displaystyle \int_{0}^{1}{{{y^2-4\,y+4}\over{2}}-{{y^2}\over{2}}\;dy}$ = 1 $\displaystyle {{\int_{0}^{1}{3\,y^2-6\,y+4\;dy}}\over{3}}$ = $\displaystyle {{2}\over{3}}$ $\displaystyle \int_{0}^{1}{\left(2-y\right)^3\;dy}$ = $\displaystyle {{15}\over{4}}$ $\displaystyle \int_{0}^{1}{\left(2-y\right)^2\;dy}$ = $\displaystyle {{7}\over{3}}$

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

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

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

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

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

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

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

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

  8. Evaluate $\displaystyle \int_{1}^{2}{\int_{1}^{4}{x^{{{3}\over{2}}}\,\left(y+2\,x\right)\;d x}\;dy}$.

    $\displaystyle \int_{1}^{2}{{{448\,y+2560}\over{35}}-{{14\,y+20}\over{35}}\;dy}$ = $\displaystyle {{3191}\over{35}}$ $\displaystyle \int_{1}^{2}{3\,\left(y+8\right)+16\;dy}$ = $\displaystyle {{89}\over{2}}$ $\displaystyle \int_{1}^{2}{8\,\left(y+8\right)-y-2\;dy}$ = $\displaystyle {{89}\over{2}}$ $\displaystyle {{5\,\int_{1}^{2}{{{320\,y+2048}\over{5}}-{{5\,y+8}\over{20}}\;dy} }\over{2}}$ = $\displaystyle {{3191}\over{35}}$

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

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

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

    $\displaystyle 2\,\int_{1}^{3}{y\,\left({{1}\over{2}}-{{y^4}\over{2}}\right)\;dy}$ = $\displaystyle -{{352}\over{3}}$ $\displaystyle \int_{-1}^{1}{y\,\left({{1}\over{2}}-{{y^4}\over{2}}\right)\;dy}$ = 0 $\displaystyle \int_{1}^{3}{y\,\left({{1}\over{2}}-{{y^4}\over{2}}\right)\;dy}$ = $\displaystyle -{{176}\over{3}}$ $\displaystyle 2\,\int_{-1}^{1}{y\,\left({{1}\over{2}}-{{y^4}\over{2}}\right)\;dy}$ = 0


Department of Mathematics
Last modified: 2026-08-11