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Calculus (Definite Integrals) - Test
∫
1
3x
2
dx
0
=
∫
2
4x
dx
0
=
∫
2
6x
2
dx
0
=
f
(x) = 6x+2
∫
4
f
(x)
dx
2
f
(x) = –4x
3
–4x
∫
1
f
(x)
dx
0
=
∫
2
–15x
2
dx
1
=
f
(x) = –24x
3
–15x
2
∫
2
f
(x)
dx
1
=
f
(x) = 15x
2
–8x–4
∫
4
f
(x)
dx
3
=
f
(x) = –21x
6
–25x
4
–36x
3
–30x
2
+3
∫
1
f
(x)
dx
0
=
f
(x) = –72x
7
+6x
5
–8x
3
∫
1
f
(x)
dx
-1
=
Učenci, ki so rešili ta test so rešili tudi :
Calculus (indefinite integrals)
Cálculo integral, integrales definidas
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