Evaluate the definite integral _ 1 ^ 2 (4 x^ 3 -5 x^ 2 +6 x+9 ) d x

Let I= _ 1 ^ 2 (4 x^ 3 -5 x^ 2 +6 x+9 ) d x I= _ 1 ^ 2 (4 x^ 3 -5 x^ 2 +6 x+9 ) d x I= _ 1 ^ 2 4 x^ 3 d x- _ 1 ^ 2 5 x^ 2 dx+ _ 1 ^ 2 6 xdx+ _ 1 ^ 2 9 dx I=4 _ 1 ^ 2 x^ 3 dx-5 _ 1 ^ 2 x^ 2 dx+6 _ 1 ^ 2 xdx+9 _ 1 ^ 2 d x I = 4 [ x^ 3+1 3+1 ]_ 1 ^ 2 -5 [ x^ 2+1 2+1 ]_ 1 ^ 2 +6 [ x^ 1+1 1+1 ]_ 1 ^ 2 +9 [ x^ 0+1 0+1 ]_ 1 ^ 2 [ x^ n d x= x^ n+1 n+1 ] I = 4 [ x^ 4 4 ]_ 1 ^ 2 -5 [ x^ 3 3 ]_ 1 ^ 2 +6 [ x^ 2 2 ]_ 1 ^ 2 +9 [x]_ 1 ^ 2 = 2^ 4 -1^ 4 -5 [ 2^ 3 3 - 1^ 3 3 ]+6 [ 2^ 2 2 - 1^ 2 2 ]+9(2-1) = 16-1-5 [7 3 ]+3(3)+9 = 33-35 3 = 99-35 3 =64 3

Evaluate the definite integral _ 1 ^ 2 (4 x^ 3 -5 x^ 2 +6 x+9 ) d…

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Question

Evaluate the definite integral $\int_{1}^{2}\left(4 x^{3}-5 x^{2}+6 x+9\right) d x$

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Answer

Let $I=\int_{1}^{2}\left(4 x^{3}-5 x^{2}+6 x+9\right) d x$
$\Rightarrow \mathrm{I}=\int_{1}^{2}\left(4 \mathrm{x}^{3}-5 \mathrm{x}^{2}+6 \mathrm{x}+9\right) \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{1}^{2} 4 \mathrm{x}^{3} \mathrm{d} \mathrm{x}-\int_{1}^{2} 5 \mathrm{x}^{2} \mathrm{dx}+\int_{1}^{2} 6 \mathrm{xdx}+\int_{1}^{2} 9 \mathrm{dx}$
$\Rightarrow \mathrm{I}=4 \int_{1}^{2} \mathrm{x}^{3} \mathrm{dx}-5 \int_{1}^{2} \mathrm{x}^{2} \mathrm{dx}+6 \int_{1}^{2} \mathrm{xdx}+9 \int_{1}^{2} \mathrm{d} \mathrm{x}$
$\Rightarrow$ I = $4 \times\left[\frac{x^{3+1}}{3+1}\right]_{1}^{2}-5 \times\left[\frac{x^{2+1}}{2+1}\right]_{1}^{2}+6 \times\left[\frac{x^{1+1}}{1+1}\right]_{1}^{2}+9 \times\left[\frac{x^{0+1}}{0+1}\right]_{1}^{2}$ [$\int x^{n} d x=\frac{x^{n+1}}{n+1}$ ]
$\Rightarrow $ I = $4 \times\left[\frac{\mathrm{x}^{4}}{4}\right]_{1}^{2}-5 \times\left[\frac{\mathrm{x}^{3}}{3}\right]_{1}^{2}+6 \times\left[\frac{\mathrm{x}^{2}}{2}\right]_{1}^{2}+9 \times[\mathrm{x}]_{1}^{2}$
= $2^{4}-1^{4}-5\left[\frac{2^{3}}{3}-\frac{1^{3}}{3}\right]+6\left[\frac{2^{2}}{2}-\frac{1^{2}}{2}\right]+9(2-1)$
= $16-1-5\left[\frac{7}{3}\right]+3(3)+9$
= $33-\frac{35}{3}$
= $\frac{99-35}{3}=\frac{64}{3}$