Definite Integrals

Question 1 of 5

Integrate

\int_{1}^{4} (2 x - 2) d x

$\int_{1}^{4} (2x-2) dx$

1. $- 9$ $-9$
2. $7$ $7$
3. $9 + c$ $9+c$
4. $9$ $9$

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Integration Formula for Definite integral

\int_{a}^{b} f (x) d x = {[F (x)]}_{a}^{b} = F (b) - F (a)

$\int_{\color{#9a00c7}{a}}^{\color{\green}{b}} f(x) dx=\left[F(x)\right]_{\color{#9a00c7}{a}}^{\color{\green}{b}}=F(\color{\green}{b})-F(\color{#9a00c7}{a})$

Integration Formula for Indefinite integral

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c

$\int x^{\color{#004ec4}{n}} dx=\frac{x^{\color{#004ec4}{n}+1}}{\color{#004ec4}{n}+1}+c$

Sum or Difference Rule

\int (f (x) \pm g (x)) d x = \int f (x) d x \pm \int g (x) d x = F (x) \pm G (x) + c

$\int (f(x) \pm g(x)) dx = \int f(x) dx \pm \int g(x) dx = F(x) \pm G(x) + c$

Find the Indefinite Integral

$\int (2x-2) dx$	$=$ $=$	$\int 2x dx-\int 2 dx$	Use the Sum or Difference Rule

	$=$ $=$	$2\int x^{\color{#004ec4}{1}} dx-2\int x^{\color{#004ec4}{0}} dx$	Take the constants out of the integral signs

	$=$ $=$	$2\frac{x^{\color{#004ec4}{1}+1}}{\color{#004ec4}{1}+1}+2\frac{x^{\color{#004ec4}{0}+1}}{\color{#004ec4}{0}+1}+c$	Use the Integration Formula for Indefinite integral

	$=$	$2x^2/2+2x^{1}/1+c$

	$=$	$x^2-2x+c$

Find the Definite Integral by using

$F(x)=x^2-2x$

$\int_{\color{#9a00c7}{1}}^{\color{\green}{4}} (2x-2) dx$	$=$	$\left[x^2-2x\right]_{\color{#9a00c7}{1}}^{\color{\green}{4}}$	Use the Integration Formula for Definite integral to get the answer

	$=$	$(\color{\green}{4}^2-2\times\color{\green}{4}) – (\color{#9a00c7}{1}^2- 2\times\color{#9a00c7}{1})$

	$=$	$(16-8) – (1-2)$

	$=$	$8 + 1$

	$=$	$9$

$9$

Question 2 of 5

Integrate

$\int_{0}^{5} 3x^{2} dx$

1. $125+c$
2. $125/3$
3. $125$
4. $-125$

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Integration Formula for Definite integral

$\int_{\color{#9a00c7}{a}}^{\color{\green}{b}} f(x) dx=\left[F(x)\right]_{\color{#9a00c7}{a}}^{\color{\green}{b}}=F(\color{\green}{b})-F(\color{#9a00c7}{a})$

Integration Formula for Indefinite integral

$\int x^{\color{#004ec4}{n}} dx=\frac{x^{\color{#004ec4}{n}+1}}{\color{#004ec4}{n}+1}+c$

Find the Indefinite Integral

$\int 3x^{2} dx$	$=$	$3\int x^{2} dx$	Take the constant $3$ out of the integral sign

	$=$	$3\frac{x^{\color{#004ec4}{2}+1}}{\color{#004ec4}{2}+1}+c$	Use the Integration Formula for Indefinite integral

	$=$	$3x^3/3+c$

	$=$	$x^3+c$

Find the Definite Integral by using

$F(x)=x^3$

$\int_{\color{#9a00c7}{0}}^{\color{\green}{5}} 3x^{2} dx$	$=$	$\left[x^3\right]_{\color{#9a00c7}{0}}^{\color{\green}{5}}$	Use the Integration Formula for Definite integral to get the answer

	$=$	$\color{\green}{5}^3 – \color{#9a00c7}{0}^3$

	$=$	$5^{3}$

	$=$	$125$

$125$

Question 3 of 5

Integrate

$\int_{-2}^{2} x^{5} dx$

1. $0$
2. $132$
3. $64$
4. $-132$

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Integration Formula for Definite integral

$\int_{\color{#9a00c7}{a}}^{\color{\green}{b}} f(x) dx=\left[F(x)\right]_{\color{#9a00c7}{a}}^{\color{\green}{b}}=F(\color{\green}{b})-F(\color{#9a00c7}{a})$

Integration Formula for Indefinite integral

$\int x^{\color{#004ec4}{n}} dx=\frac{x^{\color{#004ec4}{n}+1}}{\color{#004ec4}{n}+1}+c$

Find the Indefinite Integral

$\int x^{5} dx$	$=$	$\frac{x^{\color{#004ec4}{5}+1}}{\color{#004ec4}{5}+1}+c$	Use the Integration Formula for Indefinite integral

	$=$	$x^6/6+c$

Find the Definite Integral by using

$F(x)=x^6/6$

$\int_{\color{#9a00c7}{-2}}^{\color{\green}{2}} x^{5} dx$	$=$	$\left[\frac{x^6}{6}\right]_{\color{#9a00c7}{-2}}^{\color{\green}{2}}$	Use the Integration Formula for Definite integral to get the answer

	$=$	$\color{\green}{2}^6/6-\color{#9a00c7}{(-2)}^6/6$

	$=$	$64/6-64/6$

	$=$	$0$

$0$

Question 4 of 5

Integrate

$\int_{1}^{2} (2x-1)^{2} dx$

1. $-1 1/3$
2. $-4 1/3$
3. $1 1/3$
4. $4 1/3$

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Integration Formula for Definite integral

$\int_{\color{#9a00c7}{a}}^{\color{\green}{b}} f(x) dx=\left[F(x)\right]_{\color{#9a00c7}{a}}^{\color{\green}{b}}=F(\color{\green}{b})-F(\color{#9a00c7}{a})$

Integration Formula for Indefinite integral

$\int f(x)^{\color{#004ec4}{n}} dx=\frac{f(x)^{\color{#004ec4}{n}+1}}{(\color{#004ec4}{n}+1)f'(x)}+c$

Find the Indefinite Integral

$\int (2x-1)^{2} dx$	$=$	$\frac{(2x-1)^{\color{#004ec4}{2}+1}}{(\color{#004ec4}{2}+1)(2x-1)’}+c$	Use the Integration Formula for Indefinite integral

	$=$	$\frac{(2x-1)^{3}}{3\times2}+c$

	$=$	$\frac{(2x-1)^{3}}{6}+c$

Find the Definite Integral by using

$F(x)=(2x-1)^{3}/6$

$\int_{\color{#9a00c7}{1}}^{\color{\green}{2}} (2x-1)^{2} dx$	$=$	$\left[\frac{(2x-1)^{3}}{6}\right]_{\color{#9a00c7}{1}}^{\color{\green}{2}}$	Use the Integration Formula for Definite integral to get the answer

	$=$	$(2\times\color{\green}{2}-1)^3/6 – (2\times\color{#9a00c7}{1}-1)^3/6$

	$=$	$3^3/6 – 1/6$

	$=$	$26/6$

	$=$	$4 1/3$

$4 1/3$

Question 5 of 5

Integrate

$\int_{0}^{1} \frac{x^{3}-2x^{2}+4x}{x} dx$

1. $1/3$
2. $-3 1/3$
3. $3 1/3+c$
4. $3 1/3$

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Integration Formula for Definite integral

$\int_{\color{#9a00c7}{a}}^{\color{\green}{b}} f(x) dx=\left[F(x)\right]_{\color{#9a00c7}{a}}^{\color{\green}{b}}=F(\color{\green}{b})-F(\color{#9a00c7}{a})$

Integration Formula for Indefinite integral

$\int x^{\color{#004ec4}{n}} dx=\frac{x^{\color{#004ec4}{n}+1}}{\color{#004ec4}{n}+1}+c$

Sum or Difference Rule

$\int (f(x) \pm g(x)) dx = \int f(x) dx \pm \int g(x) dx = F(x) \pm G(x) + c$

Addition of Fractions Rule

$(a+b)/c=a/c+b/c$

Find the Indefinite Integral

$\int \frac{x^{3}-2x^{2}+4x}{x} dx$	$=$	$\int (\frac{x^{3}}{x}-\frac{2x^{2}}{x}+\frac{4x}{x}) dx$	Apply the Addition of Fractions Rule

	$=$	$\int (x^{2}-2x+4) dx$

	$=$	$\int x^{2} dx-\int 2x dx+\int 4 dx$	Use the Sum or Difference Rule

	$=$	$\int x^{2} dx-2\int x dx+4\int x^{0} dx$	Take the constants out of the integral signs

	$=$	$\frac{x^{\color{#004ec4}{2}+1}}{\color{#004ec4}{2}+1}-2\frac{x^{\color{#004ec4}{1}+1}}{\color{#004ec4}{1}+1}+4\frac{x^{\color{#004ec4}{0}+1}}{\color{#004ec4}{0}+1}+c$	Use the Integration Formula for Indefinite integral

	$=$	$x^3/3-2x^{2}/2+4x^{1}/1+c$

	$=$	$x^3/3-x^{2}+4x+c$

Find the Definite Integral by using

$F(x)=x^3/3-x^{2}+4x$

$\int_{\color{#9a00c7}{0}}^{\color{\green}{1}} \frac{x^{3}-2x^{2}+4x}{x} dx$	$=$	$\left[\frac{x^3}{3}-x^{2}+4x\right]_{\color{#9a00c7}{0}}^{\color{\green}{1}}$	Use the Integration Formula for Definite integral

	$=$	$(\color{\green}{1}^3/3-\color{\green}{1}^{2}+4\times\color{\green}{1})-(\color{#9a00c7}{0}^3/3-\color{#9a00c7}{0}^{2}+4\times\color{#9a00c7}{0})$

	$=$	$(1/3-1+4)-(0-0+0)$

	$=$	$1/3+3$

	$=$	$3 1/3$

$3 1/3$

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Quiz summary

Information

1. Question

Hint

Well Done!

Integration Formula for Definite integral

Integration Formula for Indefinite integral

Sum or Difference Rule

2. Question

Hint

Nice Job!

Integration Formula for Definite integral

Integration Formula for Indefinite integral

3. Question

Hint

Excellent!

Integration Formula for Definite integral

Integration Formula for Indefinite integral

4. Question

Hint

Keep Going!

Integration Formula for Definite integral

Integration Formula for Indefinite integral

5. Question

Hint

Correct!

Integration Formula for Definite integral

Integration Formula for Indefinite integral

Sum or Difference Rule

Addition of Fractions Rule

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