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

    $\displaystyle \int_{1}^{2}{{{y+4}\over{16}}-y-1\;dy}$ = $\displaystyle -{{7}\over{64}}$ $\displaystyle \int_{1}^{2}{{{3\,y}\over{4}}+\ln 4\;dy}$ = $\displaystyle {{4\,\ln 4+3}\over{2}}-{{8\,\ln 4+3}\over{8}}$ $\displaystyle \int_{1}^{2}{{{1}\over{16}}-{{y+4}\over{32}}\;dy}$ = $\displaystyle -{{7}\over{64}}$ $\displaystyle \int_{1}^{2}{{{y+6}\over{192}}-{{2\,y+3}\over{6}}\;dy}$ = $\displaystyle {{4\,\ln 4+3}\over{2}}-{{8\,\ln 4+3}\over{8}}$

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

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

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

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

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

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

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

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

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

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

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

    $\displaystyle \int_{-2}^{-1}{y\,\left({{9\,y^2+12\,y+4}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle -{{5}\over{8}}$ $\displaystyle \int_{0}^{1}{y\,\left({{9\,y^2+12\,y+4}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle {{97}\over{24}}$ $\displaystyle -\int_{-2}^{-1}{y\,\left({{9\,y^2+12\,y+4}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle {{5}\over{8}}$ $\displaystyle -\int_{0}^{1}{y\,\left({{9\,y^2+12\,y+4}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle -{{97}\over{24}}$

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

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

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

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

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

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

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

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

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

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


Department of Mathematics
Last modified: 2026-06-04