H a s i l d a r i ∫ sec ⁡ ( 1 − z ) tan ⁡ ( 1 − z ) d z a d a l a h . . . Hasil\ dari\int\sec\left(1-z\right)\tan\left(1-z\right)dz\ adalah\ ... H a s i l d a r i ∫ sec ( 1 − z ) tan ( 1 − z ) d z a d a l a h . . .

− sec ⁡ ( 1 − z ) + C -\sec\left(1-z\right)+C − sec ( 1 − z ) + C

sin ⁡ ( 1 − z ) + C \sin\left(1-z\right)+C sin ( 1 − z ) + C

cos ⁡ ( 1 − z ) + C \cos\left(1-z\right)+C cos ( 1 − z ) + C

sec ⁡ ( 1 − z ) + C \sec\left(1-z\right)+C sec ( 1 − z ) + C

− tan ⁡ ( 1 − z ) + C -\tan\left(1-z\right)+C − tan ( 1 − z ) + C

Kerja yang dilakukan oleh gaya F pada objek yang bergerak di sepanjang garis lurus dari a ke b adalah _____ jika F konstan, tetapi _____ jika F = F ( x ) F=F\left(x\right) F = F ( x ) merupakan variabel.

F ⋅ ( b − a ) ; ∫ a b F ( x ) d x F\cdot\left(b-a\right);\int_a^bF\left(x\right)dx F ⋅ ( b − a ) ; ∫ a b F ( x ) d x

∫ a b F d x ; ∫ a b F ( x ) d x \int_a^bFdx;\int_a^bF\left(x\right)dx ∫ a b F d x ; ∫ a b F ( x ) d x

F ⋅ ( b − a ) ; F ( x ) ⋅ ( b − a ) F\cdot\left(b-a\right);F\left(x\right)\cdot\left(b-a\right) F ⋅ ( b − a ) ; F ( x ) ⋅ ( b − a )

∫ a b F ⋅ ( b − a ) d x ; ∫ a b F ( x ) d x \int_a^bF\cdot\left(b-a\right)dx;\int_a^bF\left(x\right)dx ∫ a b F ⋅ ( b − a ) d x ; ∫ a b F ( x ) d x

∫ x n d x = \int_{ }^{ }x^{^n}dx= ∫ x n d x =

1 n + 1 x n + 1 + C \frac{1}{n+1}x^{n+1}+C n + 1 1 x n + 1 + C

1 n x n + 1 + C \frac{1}{n}x^{n+1}+C n 1 x n + 1 + C

n x n + 1 + C nx^{n+1}+C n x n + 1 + C

n x n − 1 + C nx^{n-1}+C n x n − 1 + C

∫ sin ⁡ ( 2 x ) d x = \int_{ }^{ }\sin(2x)dx= ∫ sin ( 2 x ) d x =

− 1 2 cos ⁡ ( 2 x ) + C -\frac{1}{2}\cos(2x)+C − 2 1 cos ( 2 x ) + C

1 2 cos ⁡ ( 2 x ) + C \frac{1}{2}\cos(2x)+C 2 1 cos ( 2 x ) + C

2 cos ⁡ ( 2 x ) + C 2\cos(2x)+C 2 cos ( 2 x ) + C

− 2 cos ⁡ ( 2 x ) + C -2\cos(2x)+C − 2 cos ( 2 x ) + C

∫ x n d x = \int_{ }^{ }x^{^n}dx= ∫ x n d x =

1 n + 1 x n + 1 + C \frac{1}{n+1}x^{n+1}+C n + 1 1 x n + 1 + C

1 n x n + 1 + C \frac{1}{n}x^{n+1}+C n 1 x n + 1 + C

n x n + 1 + C nx^{n+1}+C n x n + 1 + C

n x n − 1 + C nx^{n-1}+C n x n − 1 + C

∫ sin ⁡ 5 x . cos ⁡ 3 x d x \int\ \sin\ 5x\ .\ \cos\ 3x\ dx ∫ sin 5 x . cos 3 x d x

− 1 16 cos ⁡ 8 x − 1 4 cos ⁡ 2 x + C -\frac{1}{16}\ \cos\ 8x\ -\frac{1}{4}\ \cos\ 2x\ +C − 1 6 1 cos 8 x − 4 1 cos 2 x + C

1 8 sin ⁡ 8 x − 1 4 cos ⁡ 2 x \frac{1}{8}\sin\ 8x\ -\frac{1}{4}\ \cos\ 2x 8 1 sin 8 x − 4 1 cos 2 x

1 16 cos ⁡ 8 x + 1 4 sin ⁡ 2 x + C \frac{1}{16}\ \cos\ 8x\ +\ \frac{1}{4}\ \sin\ 2x\ +\ C 1 6 1 cos 8 x + 4 1 sin 2 x + C

1 4 sin ⁡ 8 x + 1 10 cos ⁡ 2 x + C \frac{1}{4}\sin\ 8x\ +\ \frac{1}{10}\ \cos\ 2x\ +\ C 4 1 sin 8 x + 1 0 1 cos 2 x + C

Jika R merupakan daerah antara kurva y = f(x) dan sumbu- x pada interval [ a, b ]. Jika f ( x ) ≥ 0 f\left(x\right)\ge0 f ( x ) ≥ 0 untuk seluruh x dalam [ a, b ], maka A ( R ) = _____, tetapi jika f ( x ) ≤ 0 f\left(x\right)\le0 f ( x ) ≤ 0 untuk seluruh x dalam [ a , b ], maka A ( R ) = ______

∫ a b f ( x ) d x ; − ∫ a b f ( x ) d x \int_a^bf\left(x\right)dx;-\int_a^bf\left(x\right)dx ∫ a b f ( x ) d x ; − ∫ a b f ( x ) d x

∫ a b f ( x ) d x ; ∫ a b f ( x ) d x \int_a^bf\left(x\right)dx;\int_a^bf\left(x\right)dx ∫ a b f ( x ) d x ; ∫ a b f ( x ) d x

∫ a b f ( x ) d x ; − ∫ b a f ( x ) d x \int_a^bf\left(x\right)dx;-\int_b^af\left(x\right)dx ∫ a b f ( x ) d x ; − ∫ b a f ( x ) d x

− ∫ a b f ( x ) d x ; − ∫ a b f ( x ) d x -\int_a^bf\left(x\right)dx;-\int_a^bf\left(x\right)dx − ∫ a b f ( x ) d x ; − ∫ a b f ( x ) d x

∫ sin ⁡ 2 x + cos ⁡ 3 x d x \int\sin\ 2x\ +\ \cos\ 3x\ dx ∫ sin 2 x + cos 3 x d x

− 1 2 cos ⁡ 2 x + 1 3 sin ⁡ 3 x + c -\frac{1}{2}\ \cos\ 2x\ \ +\ \frac{1}{3}\ \sin\ 3x\ +c − 2 1 cos 2 x + 3 1 sin 3 x + c

1 2 cos ⁡ 2 x − 1 3 sin ⁡ 3 x + C \frac{1}{2}\ \cos\ 2x\ -\frac{1}{3}\sin\ 3x\ +C 2 1 cos 2 x − 3 1 sin 3 x + C

− cos ⁡ 2 x + cos ⁡ 3 x + C -\cos\ 2x\ +\cos\ 3x\ +C − cos 2 x + cos 3 x + C

cos ⁡ 2 x − sin ⁡ 3 x + C \cos\ 2x\ -\sin\ 3x\ +C cos 2 x − sin 3 x + C

Hasil dari ∫ 0 2 π 1 − cos ⁡ x d x \int_0^{2\pi}\sqrt{1-\cos\ x}\ dx\ ∫ 0 2 π 1 − cos x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> d x adalah ....

2 2 2\sqrt{2} 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

− 2 2 -2\ \sqrt{2} − 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

− 2 -\sqrt{2} − 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 \sqrt{2} 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

∫ sin ⁡ 3 x d x \int\ \ \sin\ 3x\ dx ∫ sin 3 x d x

− 1 3 cos ⁡ 3 x + C -\frac{1}{3}\ \cos\ 3x\ \ +C − 3 1 cos 3 x + C

1 3 cos ⁡ 3 x + C \frac{1}{3}\cos\ 3x\ \ +\ C 3 1 cos 3 x + C

cos ⁡ 3 x + C \cos\ 3x\ +C cos 3 x + C

sin ⁡ 1 3 x + C \sin\ \frac{1}{3}x\ \ +\ C sin 3 1 x + C

Integral

Authored by Nesy Widyastuti

Mathematics

University

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FILL IN THE BLANKS QUESTION

30 sec • 1 pt

Nilai dari $\int_1^3\left(x^2+\frac{1}{6}\right)dx= (a)$ (tulis dalam bilangan, jika pecahan tulis a/b)

MULTIPLE SELECT QUESTION

30 sec • 1 pt

Kawat homogen terbentang pada sumbu-x antara $x=0$ dan $x=5$ akan setimbang pada $\overline{x}=$ _. Akan tetapi, jika kawat memiliki kerapatan $\delta\left(x\right)=1+x$ , titik setimbangnya berada pada $\overline{x}=$ _

$\int_0^5x\left(1+x\right)dx$

$\int_0^5\left(1+x\right)dx$

$\int_0^5xdx$

$\int_x^y\frac{1+x}{x}dx$

FILL IN THE BLANKS QUESTION

30 sec • 1 pt

Nilai dari $3\int_2^4\left(-x^2+6x-8\right)dx= (a)$ (tulis dalam bilangan, jika pecahan tulis a/b)

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\int\left(\frac{4}{5}x^3-\ 4x^2+\frac{4}{x^2}-7\right)dx\ =\ ...$

$\frac{1}{5}x^4-\frac{4}{3}x^3-4x^{-1}-7x\ +\ c$

$\frac{1}{5}x^4-\frac{4}{3}x^3+4x^{-1}-7x+c$

$\frac{1}{5}x^4-\frac{4}{3}x^3-4x^{-1}-7+c$

$\frac{1}{5}x^4-8x^3-\frac{4}{3}x^2-7x+c$

$\frac{12}{5}x^4-8x^3-4x^{-1}-7x+c$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\int_1^4(x^2-2x+2)dx=...$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

-54

-27

10.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\int_1^4(3\sqrt{x}-\frac{1}{\sqrt{x}})dx=...$

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Integral

Nilai dari ∫13(x2+16)dx=....\int_1^3\left(x^2+\frac{1}{6}\right)dx= (a) ∫13​(x2+61​)dx=.... (tulis dalam bilangan, jika pecahan tulis a/b)

Kawat homogen terbentang pada sumbu-x antara x=0x=0x=0 dan x=5x=5x=5 akan setimbang pada x‾=\overline{x}=x= _____. Akan tetapi, jika kawat memiliki kerapatan δ(x)=1+x\delta\left(x\right)=1+xδ(x)=1+x , titik setimbangnya berada pada x‾=\overline{x}=x= _____

Nilai dari 3∫24(−x2+6x−8)dx=....3\int_2^4\left(-x^2+6x-8\right)dx= (a) 3∫24​(−x2+6x−8)dx=.... (tulis dalam bilangan, jika pecahan tulis a/b)

∫(45x3− 4x2+4x2−7)dx = ...\int\left(\frac{4}{5}x^3-\ 4x^2+\frac{4}{x^2}-7\right)dx\ =\ ...∫(54​x3− 4x2+x24​−7)dx = ...

∫14(x2−2x+2)dx=...\int_1^4(x^2-2x+2)dx=...∫14​(x2−2x+2)dx=...

∫14(3x−1x)dx=...\int_1^4(3\sqrt{x}-\frac{1}{\sqrt{x}})dx=...∫14​(3x​−x​1​)dx=...

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Popular Resources on Wayground

Nilai dari $\int_1^3\left(x^2+\frac{1}{6}\right)dx= (a)$ (tulis dalam bilangan, jika pecahan tulis a/b)

Kawat homogen terbentang pada sumbu-x antara $x=0$ dan $x=5$ akan setimbang pada $\overline{x}=$ _. Akan tetapi, jika kawat memiliki kerapatan $\delta\left(x\right)=1+x$ , titik setimbangnya berada pada $\overline{x}=$ _

Nilai dari $3\int_2^4\left(-x^2+6x-8\right)dx= (a)$ (tulis dalam bilangan, jika pecahan tulis a/b)

$\int\left(\frac{4}{5}x^3-\ 4x^2+\frac{4}{x^2}-7\right)dx\ =\ ...$

$\int_1^4(x^2-2x+2)dx=...$

$\int_1^4(3\sqrt{x}-\frac{1}{\sqrt{x}})dx=...$