Question

evaluate the following definite integrals​

Answer 1

We're asked to evaluate the integral,

$\displaystyle\longrightarrow I=\int\limits_1^9\left(x^{\frac{3}{2}}+2x+3\right)\ dx$

We can distribute integral to each the three term as,

$\displaystyle\longrightarrow I=\int\limits_1^9x^{\frac{3}{2}}\ dx+\int\limits_1^92x\ dx+\int\limits_1^93\ dx$

The constants, if any, can be taken out of the integral.

$\displaystyle\longrightarrow I=\int\limits_1^9x^{\frac{3}{2}}\ dx+2\int\limits_1^9x\ dx+3\int\limits_1^9dx$

Or,

$\displaystyle\longrightarrow I=\int\limits_1^9x^{\frac{3}{2}}\ dx+2\int\limits_1^9x^1\ dx+3\int\limits_1^9x^0\ dx$

We are familiar with the formula that,

• $\displaystyle\int x^n\ dx=\dfrac{x^{n+1}}{n+1},\ n\neq-1$

By this formula, we get,

$\displaystyle\longrightarrow I=\left[\dfrac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\right]_1^9+2\left[\dfrac{x^{1+1}}{1+1}\right]_1^9+3\left[\dfrac{x^{0+1}}{0+1}\right]_1^9$

$\displaystyle\longrightarrow I=\left[\dfrac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]_1^9+2\left[\dfrac{x^2}{2}\right]_1^9+3\big[x\big]_1^9$

$\displaystyle\longrightarrow I=\dfrac{2}{5}\left[x^{\frac{5}{2}}\right]_1^9+\left[x^2\right]_1^9+3\big[x\big]_1^9$

The limit is provided as upper limit - lower limit.

$\displaystyle\longrightarrow I=\dfrac{2}{5}\left(9^{\frac{5}{2}}-1^{\frac{5}{2}}\right)+\left(9^2-1^2\right)+3\big(9-1\big)$

$\displaystyle\longrightarrow I=\dfrac{2}{5}\left(243-1\right)+\left(81-1\right)+3\times8$

$\displaystyle\longrightarrow I=\dfrac{2}{5}\times242+80+24$

$\displaystyle\longrightarrow\underline{\underline{I=200.8}}$

Hence 200.8 is the answer.