Question

Evaluate:-
$\\ \\ \int \: \bigg(25 + 32x - \frac{32}{221} \times \frac{4 + 3 + 2}{24} \times 36 - \frac{45}{3}33 \bigg)dx$
​

Answer 1

✯ To Evaluate ☟

$\\ \\ ✰ \: \: \underline{ \boxed{ \tt{ \color{purple}\int \: \bigg(25 + 32x - \frac{32}{221} \times \frac{4 + 3 + 2}{24} \times 36 - \frac{45}{3}33 \bigg) \it dx}}} \: \: ✰$

✯ Required Answer ☟

$\\\\➪\:\:\tt\displaystyle\int{ (25+32x- \frac{ 32 }{ 221 } \times \frac{ 4+3+2 }{ 24 } \times 36- \frac{ 45 }{ 3 } 33) }d x$

$➪\:\:\tt\int 25+32x-\frac{32}{221}\times \left(\frac{7+2}{24}\right)\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{32}{221}\times \left(\frac{9}{24}\right)\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{32}{221}\times \left(\frac{3}{8}\right)\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{32\times 3}{221\times 8}\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{96}{1768}\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{12}{221}\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{12\times 36}{221}-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{432}{221}-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int \frac{5525}{221}+32x-\frac{432}{221}-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int \frac{5525-432}{221}+32x-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int \frac{5093}{221}+32x-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int \frac{5093}{221}+32x-15\times 33\mathrm{d}x$

$➪\:\:\tt\int \frac{5093}{221}+32x-495\mathrm{d}x$

$➪\:\:\tt\int \frac{5093}{221}+32x-\frac{109395}{221}\mathrm{d}x$

$➪\:\:\tt\int \frac{5093-109395}{221}+32x\mathrm{d}x$

$➪\:\:\tt\int -\frac{104302}{221}+32x\mathrm{d}x$

$➪\:\:\tt\int -\frac{104302}{221}\mathrm{d}x+\int 32x\mathrm{d}x$

$➪\:\:\tt\int -\frac{104302}{221}\mathrm{d}x+32\int x\mathrm{d}x$

☛ Find the integral of 104302/221 using the table of common integrals rule $\int a\mathrm{d}x=ax.$

$➪\:\:\tt-\frac{104302x}{221}+32\int x\mathrm{d}x$

☛ Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for k$\neq$ -1, replace $\int x\mathrm{d}x$ with $\frac{x^{2}}{2}$. Multiply 32 times $\frac{x^{2}}{2}.$

$➪\:\:\tt-\frac{104302x}{221}+16x^{2}$

☛ If F$\left(x\right)$ is an antiderivative of f$\left(x\right)$, then the set of all antiderivatives of f$\left(x\right)$ is given by F$\left(x\right)$+C. Therefore, add the constant of integration C$\in \mathrm{R}$ to the result.

$✧ \: \: \underline{ \boxed{ \tt➠ \: \color{purple}-\dfrac{104302x}{221}+16x^{2}+С }} \: \: ✧$

✯ Hope it helps u ✯

Answer 2

✯ To Evaluate ☟

$\\ \\ ✰ \: \: \underline{ \boxed{ \tt{ \color{purple}\int \: \bigg(25 + 32x - \frac{32}{221} \times \frac{4 + 3 + 2}{24} \times 36 - \frac{45}{3}33 \bigg) \it dx}}} \: \: ✰$

✯ Required Answer ☟

$➪\:\:\tt\displaystyle\int{ (25+32x- \frac{ 32 }{ 221 } \times \frac{ 4+3+2 }{ 24 } \times 36- \frac{ 45 }{ 3 } 33) }d x$

$➪\:\:\tt\int 25+32x-\frac{32}{221}\times \left(\frac{7+2}{24}\right)\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{32}{221}\times \left(\frac{9}{24}\right)\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{32}{221}\times \left(\frac{3}{8}\right)\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{32\times 3}{221\times 8}\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{96}{1768}\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{12}{221}\times 36-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{12\times 36}{221}-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int 25+32x-\frac{432}{221}-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int \frac{5525}{221}+32x-\frac{432}{221}-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int \frac{5525-432}{221}+32x-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int \frac{5093}{221}+32x-\frac{45}{3}\times 33\mathrm{d}x$

$➪\:\:\tt\int \frac{5093}{221}+32x-15\times 33\mathrm{d}x$

$➪\:\:\tt\int \frac{5093}{221}+32x-495\mathrm{d}x$

$➪\:\:\tt\int \frac{5093}{221}+32x-\frac{109395}{221}\mathrm{d}x$

$➪\:\:\tt\int \frac{5093-109395}{221}+32x\mathrm{d}x$

$➪\:\:\tt\int -\frac{104302}{221}+32x\mathrm{d}x$

$➪\:\:\tt\int -\frac{104302}{221}\mathrm{d}x+\int 32x\mathrm{d}x$

$➪\:\:\tt\int -\frac{104302}{221}\mathrm{d}x+32\int x\mathrm{d}x$

Find the integral of 104302/221 using the table of common integrals rule $\int a\mathrm{d}x=ax.$

$➪\:\:\tt-\frac{104302x}{221}+32\int x\mathrm{d}x$

Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for k$\neq$ -1, replace $\int x\mathrm{d}x$ with $\frac{x^{2}}{2}$. Multiply 32 times $\frac{x^{2}}{2}.$

$➪\:\:\tt-\frac{104302x}{221}+16x^{2}$

If F$\left(x\right)$ is an antiderivative of f$\left(x\right)$, then the set of all antiderivatives of f$\left(x\right)$ is given by F$\left(x\right)$+C. Therefore, add the constant of integration C$\in \mathrm{R}$ to the result.

$✧ \: \: \underline{ \boxed{ \tt➠ \: \color{purple}-\dfrac{104302x}{221}+16x^{2}+С }} \: \: ✧$

✯ Hope it helps u ✯