Question

can someone pls solve it?​

Answer 1

Answer:

Area of shaded region=arΔABC−arΔDBC

arΔABC=

s(s−a)(s−b)(s−c)

s=

2

a+b+c

=

2

122+22+120

=132m

∴arΔABC=

132(132−122)(132−22)(132−120)

=

132×10×110×12

=

11×12×10×10×11×12

=10×11×12=1320m

2

arΔBDC=

s(s−a)(s−−b)(s−c)

s=

2

a+b+c

=

2

36+22+24

=36m

∴arΔBDC=

36(36−26)(36−22)(36−24)

=

36×10×14×12

=

12×3×2×5×2×7×12

=2×12

105

=24×10.24=245.76m

2

∴ area of shaded region=1320−245.76=1074.24m

2

≈1074m

2

Answer 2

Answer:

∫ e x d x = e x + C \int e^x \;dx = e^x+C ∫exdx=ex+C. ∫ 1 x d x = ln ⁡ x + C \int {1\over x} \;dx= \ln x+C ∫x1dx=lnx+C. ∫ sin ⁡ x d x = − cos ⁡ x + C \int \sin x\;dx=-\cos x+C ∫sinxdx=−cosx+C.In calculus, an integral is a mathematical object that can be interpreted as an area or a generalization of area.

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